Ch 7  ·  Q–
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Chapter 7 Exercise 7.7 Solutions

Integrals

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 7 Exercise 7.7

Class 12 Mathematics Exercise 7.7 NCERT Solutions Integrals Class 12 Mathematics Chapter 7 CBSE Board Exam JEE Main JEE Advanced CUET Indefinite Integration Integration of Algebraic Functions Standard Integration Formulae Integration by Substitution Completing the Square Integration of Irrational Functions Definite and Indefinite Integrals Logarithmic Integration Multiple Choice Questions Step-by-Step Solutions
11 Questions
25–35 min Ideal time
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Q1
LONG ANSWER4 marks
Integrate the following function: \[I=\int \sqrt{4-x^2}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves the integration of an algebraic function containing a square root of the form \(\sqrt{a^2-x^2}\). Such integrals can be evaluated directly using a standard integration formula derived through trigonometric substitution.

The standard integration formula is:

\[\boxed{\int \sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C}\]

Where:

  • \(a\) is a positive constant.
  • \(x\) is the variable of integration.
  • \(\sin^{-1}(x/a)\) represents the inverse sine function.
  • \(C\) is the arbitrary constant of integration.

Origin of the formula:

The formula can be derived using the trigonometric substitution:

\[ x=a\sin\theta \]

Differentiating both sides with respect to \(\theta\):

\[ dx=a\cos\theta\,d\theta \]

Also,

\[\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin^2\theta}\]

\[=a\sqrt{1-\sin^2\theta}=a\cos\theta\]

Therefore, the integral becomes:

\[ \begin{aligned} I&=\int a\cos\theta\cdot a\cos\theta\,d\theta\\ &=a^2\int\cos^2\theta\,d\theta \end{aligned} \]

Using the trigonometric identity:

\[ \cos^2\theta=\frac{1+\cos 2\theta}{2} \]

We get:

\[ \begin{aligned} I&=\frac{a^2}{2}\int(1+\cos 2\theta)\,d\theta\\ &=\frac{a^2}{2} \left(\theta+\frac{\sin 2\theta}{2}\right)+C\\ &=\frac{a^2}{2} \left(\theta+\sin\theta\cos\theta\right)+C \end{aligned} \]

Since \(x=a\sin\theta\), we have:

\[ \sin\theta=\frac{x}{a} \]

Also,

\[ \cos\theta=\frac{\sqrt{a^2-x^2}}{a} \]

Therefore,

\[ \theta=\sin^{-1}\left(\frac{x}{a}\right) \]

Substituting these values into the expression for \(I\):

\[ \begin{aligned} I&=\frac{a^2}{2} \left[ \sin^{-1}\left(\frac{x}{a}\right) + \frac{x}{a}\cdot \frac{\sqrt{a^2-x^2}}{a} \right]+C\\ &=\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) +\frac{x}{2}\sqrt{a^2-x^2}+C \end{aligned} \]

This establishes the standard integration formula used in the given problem.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Compare the expression under the square root with the standard form \(a^2-x^2\).

  2. Identify the value of the constant \(a\).

  3. Apply the standard integration formula for \(\sqrt{a^2-x^2}\).

  4. Substitute the value of \(a\) into the formula.

  5. Simplify the resulting expression and include the constant of integration.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  6 steps
  1. Given —
    \[I=\int\sqrt{4-x^2}\,dx\]
  2. Compare with the standard form.
  3. We know that:
    \[\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C\]
  4. Comparing the expression under the square root:
    \[a^2-x^2=4-x^2\]
  5. Therefore,
    \[a^2=4\implies a=2\]
  6. Apply the standard integration formula
  7. Substituting \(a=2\) into the standard formula:
    \[\begin{aligned}I&=\frac{x}{2}\sqrt{2^2-x^2}+\frac{2^2}{2}\sin^{-1}\left(\frac{x}{2}\right)+C\end{aligned}\]
  8. Simplify the expression
  9. Since \(2^2=4\), we obtain:
    \[\begin{aligned}I&=\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right)+C\\ &=\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)+C\end{aligned}\]
  10. Hence, the required integral is:
    \[\boxed{\int\sqrt{4-x^2}\,dx=\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)+C}\]
💡 Answer
Final Answer
›
\[\boxed{\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)+C}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests the ability to recognise and apply standard integration formulae.
  • It reinforces the relationship between algebraic expressions and inverse trigonometric functions.
  • It provides practice in substituting constants and simplifying mathematical expressions accurately.
  • Showing the standard formula, substitution and simplification in separate steps makes the solution easier to follow and helps demonstrate mathematical reasoning.
  • The verification by differentiation strengthens understanding of the fundamental relationship between differentiation and integration.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Recognition of standard integral forms is important for solving integration problems efficiently.
  • The formula is useful in problems involving trigonometric substitution and definite integrals.
  • Understanding the derivation helps students handle modified expressions rather than relying exclusively on memorisation.
  • The inverse trigonometric component is particularly relevant to problems involving algebraic simplification and differentiation-based verification.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Recognise the standard form: Expressions containing \(\sqrt{a^2-x^2}\) can be integrated using a standard formula.

  2. Identify the constant correctly: In this problem, \(a^2=4\), so \(a=2\).

  3. Remember the inverse trigonometric term: The integration formula contains \(\sin^{-1}(x/a)\), not \(\sin^{-1}(a/x)\).

  4. Use the positive value of \(a\): The standard formula is expressed with \(a>0\).

  5. Include the integration constant: Every indefinite integral must include the arbitrary constant \(C\).

  6. Verify whenever possible: Differentiating the final expression should reproduce the original integrand.
↑ Top
1 / 11  ·  9%
Q2 →
Q2
LONG ANSWER4 marks
Integrate the following function: \[I=\int\sqrt{1-4x^2}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves the integration of an algebraic function containing a quadratic expression under a square root. The integrand resembles the standard form \(\sqrt{a^2-u^2}\), but the variable appears as \(2x\) rather than simply \(x\).

We use the standard integration formula:

\[\boxed{\int\sqrt{a^2-u^2}\,du=\frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{u}{a}\right)+C}\]

However, since the integration is with respect to \(x\), while the expression inside the square root contains \(2x\), we must account for the differential transformation.

Let:

\[ u=2x \]

Differentiating both sides:

\[ du=2\,dx \]

Therefore:

\[ dx=\frac{du}{2} \]

This introduces a factor of \(\frac{1}{2}\) in the integral. Thus, the corresponding standard result becomes:

\[\boxed{\int\sqrt{a^2-(bx)^2}\,dx=\frac{x}{2}\sqrt{a^2-b^2x^2}+\frac{a^2}{2b}\sin^{-1}\left(\frac{bx}{a}\right)+C}\]

Here, \(a\) and \(b\) are positive constants.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rewrite \(4x^2\) as \((2x)^2\).

  2. Compare the integrand with the standard form \(\sqrt{a^2-u^2}\).

  3. Identify \(a=1\) and substitute \(u=2x\).

  4. Account for the differential transformation \(du=2dx\).

  5. Apply the integration formula and simplify the resulting expression.

  6. Verify the answer by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. Given —
    \[I=\int\sqrt{1-4x^2}\,dx\]
  2. Rewrite the expression inside the square root.
  3. Since \(4x^2=(2x)^2\), we obtain:
    \[I=\int\sqrt{1-(2x)^2}\,dx\]
  4. Introduce a suitable substitution
  5. Let:
    \[u=2x\]
  6. Differentiating with respect to \(x\):
    \[\frac{du}{dx}=2\]
  7. Therefore:
    \[du=2\,dx\]
    Or,
    \[dx=\frac{du}{2}\]
  8. Substitute into the integral
  9. Substituting \(u=2x\) and \(dx=\frac{du}{2}\):
    \[ \begin{aligned} I&=\int\sqrt{1-u^2}\cdot\frac{du}{2}\\ &=\frac{1}{2}\int\sqrt{1-u^2}\,du \end{aligned} \]
  10. Apply the standard integration formula
  11. We know that:
    \[\int\sqrt{a^2-u^2}\,du=\frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{u}{a}\right)+C\]
  12. Comparing \(\sqrt{1-u^2}\) with \(\sqrt{a^2-u^2}\), we get:
    \[a^2=1\implies a=1\]
  13. Hence:
    \[\begin{aligned}I&=\frac{1}{2}\left[\frac{u}{2}\sqrt{1-u^2}+\frac{1}{2}\sin^{-1}u\right]+C\\ &=\frac{u}{4}\sqrt{1-u^2}+\frac{1}{4}\sin^{-1}u+C\end{aligned}\]
  14. Substitute \(u=2x\) back into the result
  15. Replacing \(u\) with \(2x\):
    \[\begin{aligned}I&=\frac{2x}{4}\sqrt{1-(2x)^2}+\frac{1}{4}\sin^{-1}(2x)+C\\ &=\frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}(2x)+C\end{aligned}\]
  16. Therefore, the required integral is:
    \[\boxed{\int\sqrt{1-4x^2}\,dx=\frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}(2x)+C}\]
💡 Answer
Final Answer
›
\[\boxed{\int\sqrt{1-4x^2}\,dx=\frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}(2x)+C}\]
🎯 Exam Significance
Exam Significance
›
  • Tests the application of standard integration formulae involving quadratic expressions under square roots.
  • Develops understanding of substitution and the importance of transforming the differential correctly.
  • Reinforces the relationship between inverse trigonometric functions and integration.
  • Encourages systematic presentation of mathematical steps, including identification of constants and simplification.
  • Provides an opportunity to verify indefinite integrals through differentiation.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Tests recognition of standard integral forms with a linear expression inside the square root.
  • Strengthens substitution techniques, particularly when the coefficient of the variable is not unity.
  • Builds familiarity with inverse trigonometric functions appearing in integration results.
  • Provides a foundation for more complex integrals involving quadratic expressions and trigonometric substitutions.
  • Develops accuracy in handling numerical coefficients, an important skill in time-bound examinations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Recognise the standard form: The expression \(\sqrt{1-4x^2}\) can be rewritten as \(\sqrt{1-(2x)^2}\).

  2. Use appropriate substitution: Take \(u=2x\) to convert the integrand into the standard form.

  3. Transform the differential: Since \(du=2dx\), we must use \(dx=\frac{du}{2}\).

  4. Account for the coefficient: The substitution introduces an additional factor of \(\frac{1}{2}\), which affects both terms of the final result.

  5. Apply the chain rule: Differentiating \(\sin^{-1}(2x)\) introduces the factor \(2\).

  6. Always verify: Differentiation of the final antiderivative must reproduce the original integrand.

  7. Remember the domain: For a real-valued integrand, \(1-4x^2\geq 0\), so \(-\frac{1}{2}\leq x\leq\frac{1}{2}\).
← Q1
2 / 11  ·  18%
Q3 →
Q3
LONG ANSWER4 marks
Integrate the following function: \[I=\int\sqrt{x^2+4x+6}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves the integration of a quadratic expression under a square root. The expression \(x^2+4x+6\) is not initially in the standard form required for direct application of an integration formula.

The appropriate technique is completing the square. By expressing the quadratic polynomial as a perfect square plus a positive constant, we can transform the given integral into the standard form:

\[ \int\sqrt{x^2+a^2}\,dx \]

The standard integration formula is:

\[ \boxed{ \int\sqrt{x^2+a^2}\,dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2} \log\left|x+\sqrt{x^2+a^2}\right|+C } \]

More generally, when the expression contains a shifted variable, the formula becomes:

\[ \boxed{ \begin{aligned} \int\sqrt{(x+h)^2+a^2}\,dx ={}& \frac{x+h}{2}\sqrt{(x+h)^2+a^2}\ &+\frac{a^2}{2} \log\left|x+h+\sqrt{(x+h)^2+a^2}\right|+C \end{aligned} } \]

Here, \(h\) and \(a\) are constants. The shift \(x+h\) is handled by substitution, while the logarithmic term arises from integrating the square-root expression.

Concept of completing the square:

For a quadratic expression of the form:

\[ x^2+bx+c \]

We use the identity:

\[ x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c-\frac{b^2}{4}\right) \]

This transformation is particularly useful when the resulting constant is positive, because it produces an expression of the form \(\sqrt{u^2+a^2}\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given quadratic expression under the square root.

  2. Complete the square by adding and subtracting the appropriate constant.

  3. Express the resulting quadratic as a perfect square plus a positive constant.

  4. Introduce a substitution to convert the shifted variable into a standard variable.

  5. Apply the standard integration formula for \(\sqrt{u^2+a^2}\).

  6. Substitute the original variable back and simplify the final expression.

  7. Verify the result by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Given —
    \[I=\int\sqrt{x^2+4x+6}\,dx\]
  2. Complete the square
  3. Consider the quadratic expression:
    \[x^2+4x+6\]
  4. To complete the square, take half the coefficient of \(x\) and square it.
  5. The coefficient of \(x\) is \(4\).
  6. Therefore:
    \[\left(\frac{4}{2}\right)^2=2^2=4\]
  7. Add and subtract \(4\) within the expression:
    \[ \begin{aligned} x^2+4x+6 &=x^2+4x+4-4+6\\ &=(x^2+4x+4)+2\\ &=(x+2)^2+2 \end{aligned} \]
  8. Since \(2=(\sqrt{2})^2\), we can write:
    \[\boxed{x^2+4x+6=(x+2)^2+(\sqrt{2})^2}\]
  9. Rewrite the integral in standard form.
  10. Substituting the completed-square expression into the integral:
    \[I=\int\sqrt{(x+2)^2+(\sqrt{2})^2}\,dx\]
  11. Comparing this with the standard form \(\sqrt{u^2+a^2}\), we identify:
    \[u=x+2,\quad a=\sqrt{2}\]
  12. Introduce substitution
  13. Let
    \[u=x+2\]
  14. Differentiating both sides:
    \[du=dx\]
  15. Also:
    \[a^2=(\sqrt{2})^2=2\]
  16. Therefore, the integral becomes:
    \[I=\int\sqrt{u^2+2}\,du\]
  17. Apply the standard integration formula
  18. We know that:
    \[\int\sqrt{u^2+a^2}\,du=\frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\log\left|u+\sqrt{u^2+a^2}\right|+C\]
  19. Substituting \(a^2=2\):
    \[\begin{aligned}I&=\frac{u}{2}\sqrt{u^2+2}+\frac{2}{2}\log\left|u+\sqrt{u^2+2}\right|+C\\ &=\frac{u}{2}\sqrt{u^2+2}+\log\left|u+\sqrt{u^2+2}\right|+C\end{aligned}\]
  20. Substitute \(u=x+2\) back into the expression.
  21. Replacing \(u\) with \(x+2\):
    \[\begin{aligned}I={}&\frac{x+2}{2}\sqrt{(x+2)^2+2}\\ &+\log\left|x+2+\sqrt{(x+2)^2+2}\right|+C\end{aligned}\]
  22. Since:
    \[(x+2)^2+2=x^2+4x+6\]
  23. The final expression becomes:
    \[\boxed{\begin{aligned}\int\sqrt{x^2+4x+6}\,dx={}&\frac{x+2}{2}\sqrt{x^2+4x+6}\\ &+\log\left|x+2+\sqrt{x^2+4x+6}\right|+C\end{aligned}}\]
💡 Answer
Final Answer
›
\[\boxed{\begin{aligned}\int\sqrt{x^2+4x+6}\,dx={}&\frac{x+2}{2}\sqrt{x^2+4x+6}\\ &+\log\left|x+2+\sqrt{x^2+4x+6}\right|+C\end{aligned}}\]
🎯 Exam Significance
Exam Significance
›

For CBSE Board Examination Aspirants:

  • Completing the square: Tests the ability to transform a quadratic polynomial into a standard algebraic form.
  • Application of standard formulae: Reinforces the use of integration formulae involving square roots of quadratic expressions.
  • Substitution technique: Develops the ability to handle shifted variables without introducing unnecessary complications.
  • Logarithmic functions: Strengthens understanding of logarithmic terms appearing in indefinite integrals.
  • Mathematical presentation: Demonstrates the importance of showing each algebraic transformation clearly and systematically.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Completing the square is an important technique for simplifying integrals involving quadratic polynomials.
  • Recognition of the standard form \(\sqrt{x^2+a^2}\) helps improve problem-solving speed.
  • Understanding the logarithmic component is useful in problems involving inverse hyperbolic functions and logarithmic transformations.
  • The substitution technique is frequently useful in more complex integration problems involving shifted quadratic expressions.
  • Verification by differentiation develops accuracy and helps identify errors in algebraic manipulation.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Completing the square: Convert \(x^2+4x+6\) into \((x+2)^2+2\).

  2. Recognising the standard form: The transformed expression belongs to the form \(\sqrt{u^2+a^2}\).

  3. Appropriate substitution: Take \(u=x+2\), which gives \(du=dx\).

  4. Identify the constant correctly: Here, \(a^2=2\), not \(a=2\).

  5. Remember the logarithmic coefficient: The coefficient of the logarithmic term is \(\frac{a^2}{2}=\frac{2}{2}=1\).

  6. Maintain the correct quadratic expression: After substitution, \((x+2)^2+2\) must simplify back to \(x^2+4x+6\).

  7. Include the constant of integration: Always add \(C\) for an indefinite integral.

  8. Verify the answer: Differentiating the final expression must reproduce the original integrand.
← Q2
3 / 11  ·  27%
Q4 →
Q4
LONG ANSWER4 marks
Integrate the following function: \[I=\int\sqrt{x^2+4x+1}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves the integration of an algebraic function containing the square root of a quadratic expression. The appropriate technique is completing the square, followed by applying the standard integration formula for the square root of the difference of two squares.

The given quadratic expression can be transformed into:

\[ x^2+4x+1=(x+2)^2-3 \]

Consequently, the integral takes the standard form:

\[ \int\sqrt{u^2-a^2}\,du \]

Standard Integration Formula:

\[\boxed{\int\sqrt{x^2-a^2}\,dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C}\]

Here, \(a\) is a positive constant and \(C\) is the arbitrary constant of integration.

When the expression contains a shifted variable, the corresponding formula is:

\[\boxed{\begin{aligned}\int\sqrt{(x+h)^2-a^2}\,dx={}&\frac{x+h}{2}\sqrt{(x+h)^2-a^2}\\ &-\frac{a^2}{2}\log\left|x+h+\sqrt{(x+h)^2-a^2}\right|+C\end{aligned}}\]

This formula is applicable on intervals where \((x+h)^2-a^2\geq 0\). It is derived using substitution and the standard integration techniques for algebraic functions.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the quadratic expression under the square root.

  2. Complete the square by adding and subtracting the appropriate constant.

  3. Express the quadratic as the difference of two squares.

  4. Introduce a substitution to eliminate the shift in the variable.

  5. Identify the constants \(a\) and \(a^2\) in the standard integration formula.

  6. Apply the formula and substitute the original variable back.

  7. Simplify the expression and verify the result by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Given —
    \[I=\int\sqrt{x^2+4x+1}\,dx\]
  2. Complete the square
  3. Consider the quadratic expression:
    \[x^2+4x+1\]
  4. The coefficient of \(x\) is \(4\). Half of this coefficient is:
    \[\frac{4}{2}=2\]
  5. Squaring this value:
    \[2^2=4\]
  6. Add and subtract \(4\) to complete the square:
    \[ \begin{aligned} x^2+4x+1 &=x^2+4x+4-4+1\\ &=(x^2+4x+4)-3\\ &=(x+2)^2-3 \end{aligned} \]
  7. Since \(3=(\sqrt{3})^2\), we can write:
    \[\boxed{x^2+4x+1=(x+2)^2-(\sqrt{3})^2}\]
  8. Rewrite the integral in standard form
  9. Substituting the completed-square expression:
    \[I=\int\sqrt{(x+2)^2-(\sqrt{3})^2}\,dx\]
  10. Comparing this expression with the standard form \(\sqrt{u^2-a^2}\), we identify:
    \[u=x+2\]
    And:
    \[a=\sqrt{3},\qquad a^2=3\]
  11. Introduce substitution
  12. Let
    \[u=x+2\]
  13. Differentiating both sides:
    \[du=dx\]
  14. Therefore, the integral becomes:
    \[I=\int\sqrt{u^2-3}\,du\]
  15. Apply the standard integration formula
  16. We know that:
    \[\int\sqrt{u^2-a^2}\,du=\frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\log\left|u+\sqrt{u^2-a^2}\right|+C\]
  17. Substituting \(a^2=3\):
    \[ \begin{aligned} I &= \frac{u}{2}\sqrt{u^2-3} - \frac{3}{2} \log\left|u+\sqrt{u^2-3}\right|+C \end{aligned} \]
  18. Substitute \(u=x+2\) back into the expression.
  19. Replacing \(u\) with \(x+2\):
    \[ \begin{aligned} I={}& \frac{x+2}{2}\sqrt{(x+2)^2-3}\\ &-\frac{3}{2} \log\left|x+2+\sqrt{(x+2)^2-3}\right|+C \end{aligned} \]
  20. Since:
    \[(x+2)^2-3=x^2+4x+1\]
  21. We obtain:
    \[ \boxed{ \begin{aligned} \int\sqrt{x^2+4x+1}\,dx ={}& \frac{x+2}{2}\sqrt{x^2+4x+1}\\ &-\frac{3}{2} \log\left|x+2+\sqrt{x^2+4x+1}\right|+C \end{aligned} } \]
💡 Answer
Final Answer
›
🎯 Exam Significance
Exam Significance
›
  • Completing the square: Tests the ability to transform a quadratic expression into a standard algebraic form.
  • Application of standard formulae: Reinforces the integration formula involving \(\sqrt{x^2-a^2}\).
  • Substitution technique: Develops the ability to handle shifted variables systematically.
  • Logarithmic integration: Strengthens understanding of logarithmic terms arising in integration.
  • Step-by-step presentation: Demonstrates how algebraic transformations and standard formulae should be applied logically.
  • Verification: Differentiation provides an independent method of checking the final answer.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Tests the recognition of integrals involving the difference of two squares under a radical.
  • Strengthens the technique of completing the square, which is useful in several forms of algebraic integration.
  • Develops familiarity with logarithmic expressions generated by standard integration formulae.
  • Builds the foundation for solving more complex integrals involving quadratic expressions and substitutions.
  • Encourages precision in handling negative signs and coefficients, especially in logarithmic terms.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Complete the square: Transform \(x^2+4x+1\) into \((x+2)^2-3\).

  2. Recognise the standard form: The resulting expression belongs to \(\sqrt{u^2-a^2}\).

  3. Use an appropriate substitution: Take \(u=x+2\), which gives \(du=dx\).

  4. Identify the constant correctly: Here, \(a=\sqrt{3}\) and \(a^2=3\).

  5. Remember the negative logarithmic term: For \(\sqrt{x^2-a^2}\), the standard formula contains \(-\frac{a^2}{2}\log|\cdots|\).

  6. Maintain algebraic accuracy: The expression \((x+2)^2-3\) must simplify to \(x^2+4x+1\).

  7. Check the domain: For a real-valued integrand, \(x^2+4x+1\geq0\), which gives \(x\leq-2-\sqrt{3}\) or \(x\geq-2+\sqrt{3}\).

  8. Verify by differentiation: The derivative of the final expression must equal the original integrand.
← Q3
4 / 11  ·  36%
Q5 →
Q5
LONG ANSWER4 marks
Integrate the following function \[I=\int\sqrt{1-4x-x^2}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves the integration of a quadratic expression under a square root. The expression \(1-4x-x^2\) is not initially in the standard form required for direct application of an integration formula.

We can simplify it by using the technique of completing the square. After rearranging and completing the square, the given quadratic expression can be written as:

\[ 1-4x-x^2=5-(x+2)^2 \]

Consequently, the integral takes the standard form:

\[ \int\sqrt{a^2-u^2}\,du \]

Standard Integration Formula:

\[ \boxed{ \int\sqrt{a^2-u^2}\,du = \frac{u}{2}\sqrt{a^2-u^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{u}{a}\right)+C } \]

Here:

  • \(a\) is a positive constant.
  • \(u\) is the variable of integration.
  • \(\sin^{-1}(u/a)\) represents the inverse sine function.
  • \(C\) is the arbitrary constant of integration.

When the expression contains a shifted variable \(x+h\), the corresponding formula becomes:

\[ \boxed{ \begin{aligned} \int\sqrt{a^2-(x+h)^2}\,dx ={}& \frac{x+h}{2}\sqrt{a^2-(x+h)^2}\\[6pt] &+\frac{a^2}{2} \sin^{-1}\left(\frac{x+h}{a}\right)+C \end{aligned} } \]

This formula is applicable when the expression under the square root is non-negative. It is obtained by using a trigonometric substitution of the form \(u=a\sin\theta\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rearrange the quadratic expression to make the coefficient of \(x^2\) positive inside the bracket.

  2. Complete the square by adding and subtracting the appropriate constant.

  3. Express the resulting quadratic as the difference of two squares.

  4. Identify the shifted variable and the constant in the standard integration formula.

  5. Apply the standard formula for \(\sqrt{a^2-u^2}\).

  6. Substitute the original variable back and simplify the expression.

  7. Verify the final result by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Given —
    \[I=\int\sqrt{1-4x-x^2}\,dx\]
  2. Rearrange the quadratic expression
  3. Consider the expression under the square root:
    \[1-4x-x^2\]
  4. Rearranging the terms:
    \[=-x^2-4x+1\]
  5. Factor out \(-1\) from the quadratic terms:
    \[=-(x^2+4x)+1\]
  6. Complete the square
  7. The coefficient of \(x\) inside the bracket is \(4\). Half of this coefficient is:
    \[\frac{4}{2}=2\]
  8. Squaring this value:
    \[2^2=4\]
  9. Add and subtract \(4\) inside the quadratic expression:
    \[ \begin{aligned} 1-4x-x^2 &=1-(x^2+4x+4)+4\\ &=5-(x^2+4x+4)\\ &=5-(x+2)^2 \end{aligned} \]
  10. Since \(5=(\sqrt{5})^2\), we obtain:
    \[1-4x-x^2=(\sqrt{5})^2-(x+2)^2\]
  11. Rewrite the integral in standard form.
  12. Substituting the completed-square expression into the integral:
    \[I=\int\sqrt{(\sqrt{5})^2-(x+2)^2}\,dx\]
  13. Comparing this with the standard form \(\sqrt{a^2-u^2}\), we identify:
    \[u=x+2\]
    and
    \[a=\sqrt{5},\qquad a^2=5\]
  14. Introduce substitution
  15. Let
    \[u=x+2\]
  16. Differentiating both sides:
    \[du=dx\]
  17. Therefore, the integral becomes:
    \[I=\int\sqrt{5-u^2}\,du\]
  18. Apply the standard integration formula
  19. We know that:
    \[\int\sqrt{a^2-u^2}\,du=\frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{u}{a}\right)+C\]
  20. Substituting \(a^2=5\):
    \[\begin{aligned}I&=\frac{u}{2}\sqrt{5-u^2}+\frac{5}{2}\sin^{-1}\left(\frac{u}{\sqrt{5}}\right)+C\end{aligned}\]
  21. Substitute \(u=x+2\) back into the expression
  22. Replacing \(u\) with \(x+2\):
    \[ \begin{aligned} I={}& \frac{x+2}{2}\sqrt{5-(x+2)^2}\\ &+\frac{5}{2} \sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C \end{aligned} \]
  23. Since:
    \[5-(x+2)^2=1-4x-x^2\]
  24. The final expression becomes:
    \[ \boxed{ \begin{aligned} \int\sqrt{1-4x-x^2}\,dx ={}& \frac{x+2}{2}\sqrt{1-4x-x^2}\\ &+\frac{5}{2} \sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C \end{aligned} } \]
💡 Answer
Final Answer
›
\[ \boxed{ \begin{aligned} \int\sqrt{1-4x-x^2}\,dx ={}& \frac{x+2}{2}\sqrt{1-4x-x^2}\\ &+\frac{5}{2} \sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C \end{aligned} } \]
🎯 Exam Significance
Exam Significance
›
  • Completing the square: Tests the ability to transform a quadratic expression into a standard algebraic form.
  • Application of standard formulae: Reinforces the integration formula involving \(\sqrt{a^2-x^2}\).
  • Substitution technique: Develops the ability to handle shifted variables systematically.
  • Inverse trigonometric functions: Strengthens the connection between algebraic integration and inverse trigonometric expressions.
  • Step-by-step mathematical presentation: Demonstrates the importance of showing algebraic transformations and substitutions clearly.
  • Verification: Encourages the use of differentiation to check the correctness of an indefinite integral.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Tests recognition of integrals involving the difference of two squares under a radical.
  • Develops proficiency in completing the square, a useful technique in several integration problems.
  • Reinforces the application of inverse trigonometric integration formulae.
  • Builds a foundation for solving more complex integrals involving quadratic expressions and trigonometric substitutions.
  • Improves accuracy in handling shifted variables and coefficients in inverse trigonometric functions.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Complete the square: Transform \(1-4x-x^2\) into \(5-(x+2)^2\).

  2. Recognise the standard form: The transformed expression belongs to \(\sqrt{a^2-u^2}\).

  3. Use appropriate substitution: Take \(u=x+2\), which gives \(du=dx\).

  4. Identify the constant correctly: Here, \(a=\sqrt{5}\) and \(a^2=5\).

  5. Remember the inverse sine term: The formula contains \(\sin^{-1}(u/a)\).

  6. Maintain algebraic accuracy: The expression \(5-(x+2)^2\) must simplify to \(1-4x-x^2\).

  7. Check the real-valued domain: The integrand is real when \(1-4x-x^2\geq0\), which gives \(-2-\sqrt{5}\leq x\leq-2+\sqrt{5}\).

  8. Verify by differentiation: Differentiating the final expression must reproduce the original integrand.
← Q4
5 / 11  ·  45%
Q6 →
Q6
LONG ANSWER4 marks
Integrate the following function: \[I=\int\sqrt{x^2+4x-5}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves the integration of the square root of a quadratic expression. The given expression can be transformed into the standard form \(\sqrt{u^2-a^2}\) by completing the square.

The relevant standard integration formula is:

\[\boxed{\int\sqrt{u^2-a^2}\,du=\frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\ln\left|u+\sqrt{u^2-a^2}\right|+C}\]

Here, \(a\) is a positive constant and \(u\) is the variable of integration. The absolute value is important because the logarithm must have a positive argument wherever the real-valued integral is defined.

Since the quadratic expression contains a linear term, we first complete the square and then use a suitable substitution.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the quadratic expression under the square root.

  2. Complete the square to express it as a difference of two squares.

  3. Introduce a substitution to obtain the standard form.

  4. Apply the integration formula for \(\sqrt{u^2-a^2}\).

  5. Substitute the original variable back into the result.

  6. Verify the answer by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Given —
    \[I=\int\sqrt{x^2+4x-5}\,dx\]
  2. Complete the square
  3. Consider the quadratic expression:
    \[x^2+4x-5\]
  4. To complete the square, take half the coefficient of \(x\) and square it:
    \[\left(\frac{4}{2}\right)^2=2^2=4\]
  5. Add and subtract \(4\):
    \[ \begin{aligned} x^2+4x-5 &=x^2+4x+4-4-5\\ &=(x^2+4x+4)-9\\ &=(x+2)^2-9\\ &=(x+2)^2-3^2 \end{aligned} \]
  6. Therefore:
    \[\boxed{x^2+4x-5=(x+2)^2-3^2}\]
  7. Rewrite the integral
  8. Substituting the completed-square expression:
    \[I=\int\sqrt{(x+2)^2-3^2}\,dx\]
  9. Apply substitution
  10. Let:
    \[u=x+2\]
  11. Differentiating both sides:
    \[du=dx\]
    Also, \(a=3\). Therefore, the integral becomes:
    \[I=\int\sqrt{u^2-3^2}\,du\]
  12. Apply the standard integration formula
  13. Recall the formula:
    \[ \int\sqrt{u^2-a^2}\,du = \frac{u}{2}\sqrt{u^2-a^2} - \frac{a^2}{2} \ln\left|u+\sqrt{u^2-a^2}\right|+C \]
  14. Substituting \(a=3\), so that \(a^2=9\):
    \[ \begin{aligned} I={}& \frac{u}{2}\sqrt{u^2-9}\\ &-\frac{9}{2} \ln\left|u+\sqrt{u^2-9}\right|+C \end{aligned} \]
  15. Substitute \(u=x+2\) back
  16. Replacing \(u\) with \(x+2\):
    \[ \begin{aligned} I={}& \frac{x+2}{2}\sqrt{(x+2)^2-9}\\ &-\frac{9}{2} \ln\left|x+2+\sqrt{(x+2)^2-9}\right|+C \end{aligned} \]
  17. Since:
    \[(x+2)^2-9=x^2+4x-5\]
  18. The final result is:
    \[ \boxed{ \begin{aligned} \int\sqrt{x^2+4x-5}\,dx ={}& \frac{x+2}{2}\sqrt{x^2+4x-5}\\ &-\frac{9}{2} \ln\left|x+2+\sqrt{x^2+4x-5}\right|+C \end{aligned} } \]
💡 Answer
Final Answer
›
\[ \boxed{ \begin{aligned} \int\sqrt{x^2+4x-5}\,dx ={}& \frac{x+2}{2}\sqrt{x^2+4x-5}\\ &-\frac{9}{2} \ln\left|x+2+\sqrt{x^2+4x-5}\right|+C \end{aligned} } \]
🎯 Exam Significance
Exam Significance
›
  • Completing the square: Demonstrates how to transform a quadratic expression into a standard algebraic form.
  • Standard integration formula: Reinforces the application of the formula involving \(\sqrt{x^2-a^2}\).
  • Substitution: Develops the ability to handle expressions containing shifted variables.
  • Logarithmic functions: Strengthens understanding of logarithmic terms arising from algebraic integration.
  • Mathematical verification: Demonstrates how differentiation can be used to confirm an indefinite integral.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Tests recognition of the standard form \(\sqrt{u^2-a^2}\).
  • Develops proficiency in completing the square and making appropriate substitutions.
  • Reinforces the relationship between algebraic expressions and logarithmic functions in integration.
  • Builds a foundation for more advanced integrals involving quadratic expressions under radicals.
  • Encourages careful handling of logarithmic arguments and algebraic simplification.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Complete the square: Transform \(x^2+4x-5\) into \((x+2)^2-9\).

  2. Recognise the standard form: The transformed expression is of the form \(\sqrt{u^2-a^2}\).

  3. Choose the correct substitution: Take \(u=x+2\), giving \(du=dx\).

  4. Identify the constant: Here, \(a=3\) and \(a^2=9\).

  5. Remember the negative logarithmic term: The formula for \(\sqrt{u^2-a^2}\) contains \(-\frac{a^2}{2}\ln|u+\sqrt{u^2-a^2}|\).

  6. Maintain the correct radical expression: The logarithmic term must contain the same radical \(\sqrt{x^2+4x-5}\).

  7. Check the real-valued domain: The radicand is non-negative when \(x\leq-5\) or \(x\geq1\).

  8. Verify the result: Differentiation of the final expression must reproduce the original integrand.
← Q5
6 / 11  ·  55%
Q7 →
Q7
LONG ANSWER4 marks
Integrate the following function: \[I=\int\sqrt{1+3x-x^2}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves integrating the square root of a quadratic expression. The appropriate technique is completing the square, which transforms the quadratic expression into the standard form:

\[ \sqrt{a^2-u^2} \]

We use the standard integration formula:

\[ \boxed{ \int\sqrt{a^2-u^2}\,du = \frac{u}{2}\sqrt{a^2-u^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{u}{a}\right)+C } \]

The formula is particularly useful when the quadratic expression has a negative coefficient of \(x^2\). Completing the square allows us to identify the shifted variable and the positive constant required for the standard formula.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rearrange the quadratic expression to make completing the square easier.

  2. Complete the square for the quadratic terms.

  3. Express the result as the difference of two squares.

  4. Introduce a suitable substitution.

  5. Apply the standard integration formula for \(\sqrt{a^2-u^2}\).

  6. Substitute the original variable back into the result.

  7. Verify the answer by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Given —
    \[I=\int\sqrt{1+3x-x^2}\,dx\]
  2. Rearrange the quadratic expression
  3. Consider the expression under the square root:
    \[1+3x-x^2\]
  4. Rearranging:
    \[=-(x^2-3x-1)\]
  5. Complete the square
  6. The coefficient of \(x\) inside the bracket is \(-3\). Half of this coefficient is:
    \[\frac{-3}{2}=-\frac32\]
  7. Squaring this value:
    \[\left(-\frac32\right)^2=\frac94\]
  8. Add and subtract \(\frac94\):
  9. \[ \begin{aligned} 1+3x-x^2 &=-\left(x^2-3x-1\right)\\ &=-\left(x^2-3x+\frac94-\frac94-1\right)\\ &=-\left[ \left(x-\frac32\right)^2-\frac{13}{4} \right]\\ &=\frac{13}{4}-\left(x-\frac32\right)^2 \end{aligned} \]
  10. Therefore:
    \[\boxed{1+3x-x^2=\left(\frac{\sqrt{13}}{2}\right)^2-\left(x-\frac32\right)^2}\]
  11. Rewrite the integral in standard form
  12. Substituting the completed-square expression:
    \[I=\int\sqrt{\left(\frac{\sqrt{13}}{2}\right)^2-\left(x-\frac32\right)^2}\,dx\]
  13. Introduce substitution
  14. Let
    \[u=x-\frac32\]
  15. Differentiating both sides:
    \[du=dx\]
  16. Also, let:
    \[a=\frac{\sqrt{13}}{2}\quad\Rightarrow\quad a^2=\frac{13}{4}\]
  17. The integral becomes:
    \[I=\int\sqrt{\frac{13}{4}-u^2}\,du\]
  18. Apply the standard integration formula
  19. Using:
    \[\int\sqrt{a^2-u^2}\,du=\frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{u}{a}\right)+C\]
  20. Substituting \(a^2=\frac{13}{4}\):
    \[ \begin{aligned} I={}& \frac{u}{2}\sqrt{\frac{13}{4}-u^2}\\ &+\frac{13}{8} \sin^{-1}\left(\frac{2u}{\sqrt{13}}\right)+C \end{aligned} \]
  21. Substitute \(u=x-\frac32\) back
  22. Replacing \(u\) with \(x-\frac32\):
    \[ \begin{aligned} I={}& \frac{x-\frac32}{2} \sqrt{\frac{13}{4}-\left(x-\frac32\right)^2}\\ &+\frac{13}{8} \sin^{-1}\left( \frac{2x-3}{\sqrt{13}} \right)+C \end{aligned} \]
  23. Since:
    \[ \frac{13}{4}-\left(x-\frac32\right)^2 =1+3x-x^2 \]
    and
    \[\frac{x-\frac32}{2}=\frac{2x-3}{4}\]
  24. The final answer is:
    \[ \boxed{ \begin{aligned} \int\sqrt{1+3x-x^2}\,dx ={}& \frac{2x-3}{4}\sqrt{1+3x-x^2}\\ &+\frac{13}{8} \sin^{-1}\left( \frac{2x-3}{\sqrt{13}} \right)+C \end{aligned} } \]
💡 Answer
Final Answer
›
\[ \boxed{ \begin{aligned} \int\sqrt{1+3x-x^2}\,dx ={}& \frac{2x-3}{4}\sqrt{1+3x-x^2}\\ &+\frac{13}{8} \sin^{-1}\left( \frac{2x-3}{\sqrt{13}} \right)+C \end{aligned} } \]
🎯 Exam Significance
Exam Significance
›
  • Completing the square: Tests the ability to transform a quadratic expression into a standard form.
  • Standard integration formula: Reinforces the formula involving \(\sqrt{a^2-x^2}\).
  • Substitution technique: Develops accuracy in handling shifted variables.
  • Inverse trigonometric functions: Strengthens the connection between algebraic integration and inverse trigonometric expressions.
  • Presentation: Encourages clear algebraic transformations and systematic application of formulae.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Tests recognition of integrals involving a quadratic expression under a square root.
  • Develops fluency in completing the square and applying substitutions.
  • Reinforces the use of inverse trigonometric functions in integration.
  • Builds a foundation for more advanced problems involving quadratic radicals.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Complete the square: \(1+3x-x^2=\frac{13}{4}-(x-\frac32)^2\).

  2. Recognise the standard form: The expression is of the form \(\sqrt{a^2-u^2}\).

  3. Use substitution: Take \(u=x-\frac32\).

  4. Identify the constant: \(a=\frac{\sqrt{13}}{2}\) and \(a^2=\frac{13}{4}\).

  5. Apply the inverse sine formula: The inverse sine argument is \(\frac{2x-3}{\sqrt{13}}\).

  6. Check the domain: The integrand is real for \(\frac{3-\sqrt{13}}{2}\leq x\leq\frac{3+\sqrt{13}}{2}\).
← Q6
7 / 11  ·  64%
Q8 →
Q8
LONG ANSWER4 marks
Integrate the following function: \[I=\int\sqrt{x^2+3x}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves the integration of the square root of a quadratic expression. The expression \(x^2+3x\) can be transformed into the standard form \(\sqrt{u^2-a^2}\) by completing the square.

The relevant standard integration formula is:

\[\boxed{\int\sqrt{u^2-a^2}\,du=\frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\ln\left|u+\sqrt{u^2-a^2}\right|+C}\]

The main idea is to complete the square, identify the shifted variable, and then apply this formula.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the quadratic expression under the square root.

  2. Complete the square by adding and subtracting the appropriate constant.

  3. Express the quadratic as the difference of two squares.

  4. Introduce a suitable substitution.

  5. Apply the standard integration formula for \(\sqrt{u^2-a^2}\).

  6. Substitute the original variable back into the result.

  7. Verify the answer by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Given —
    \[I=\int\sqrt{x^2+3x}\,dx\]
  2. Complete the square
  3. Consider the quadratic expression:
    \[x^2+3x\]
  4. The coefficient of \(x\) is \(3\). Half of this coefficient is:
    \[\frac{3}{2}\]
  5. Squaring this value:
    \[\left(\frac32\right)^2=\frac94\]
  6. Add and subtract \(\frac94\):
    \[ \begin{aligned} x^2+3x &=x^2+3x+\frac94-\frac94\\ &=\left(x+\frac32\right)^2-\frac94\\ &=\left(x+\frac32\right)^2-\left(\frac32\right)^2 \end{aligned} \]
  7. Therefore:
    \[\boxed{x^2+3x=\left(x+\frac32\right)^2-\left(\frac32\right)^2}\]
  8. Rewrite the integral
  9. Substituting the completed-square expression:
    \[I=\int\sqrt{\left(x+\frac32\right)^2-\left(\frac32\right)^2}\,dx\]
  10. Introduce substitution
  11. Let:
    \[u=x+\frac32\]
  12. Differentiating both sides:
    \[du=dx\]
  13. Also:
    \[a=\frac32,\quad a^2=\frac94\]
  14. Hence, the integral becomes:
    \[I=\int\sqrt{u^2-\frac94}\,du\]
  15. Apply the standard integration formula
  16. Recall:
    \[\int\sqrt{u^2-a^2}\,du=\frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\ln\left|u+\sqrt{u^2-a^2}\right|+C\]
  17. Substituting \(a^2=\frac94\):
    \[ \begin{aligned} I={}& \frac{u}{2}\sqrt{u^2-\frac94}\\ &-\frac98 \ln\left|u+\sqrt{u^2-\frac94}\right|+C \end{aligned} \]
  18. Substitute \(u=x+\frac32\) back
  19. Replacing \(u\) with \(x+\frac32\):
    \[ \begin{aligned} I={}& \frac{x+\frac32}{2} \sqrt{\left(x+\frac32\right)^2-\frac94}\\ &-\frac98 \ln\left| x+\frac32+ \sqrt{\left(x+\frac32\right)^2-\frac94} \right|+C \end{aligned} \]
  20. Since:
    \[\left(x+\frac32\right)^2-\frac94=x^2+3x\]
    And:
    \[\frac{x+\frac32}{2}=\frac{2x+3}{4}\]
  21. The final result is:
    \[ \boxed{ \begin{aligned} \int\sqrt{x^2+3x}\,dx ={}& \frac{2x+3}{4}\sqrt{x^2+3x}\\ &-\frac98 \ln\left| x+\frac32+\sqrt{x^2+3x} \right|+C \end{aligned} } \]
💡 Answer
Final Answer
›
\[\boxed{\begin{aligned}\int\sqrt{x^2+3x}\,dx={}&\frac{2x+3}{4}\sqrt{x^2+3x}\\ &-\frac98\ln\left|x+\frac32+\sqrt{x^2+3x}\right|+C\end{aligned}}\]
🎯 Exam Significance
Exam Significance
›
  • Completing the square: Demonstrates how to transform a quadratic expression into a standard algebraic form.
  • Standard integration formula: Reinforces the formula involving \(\sqrt{x^2-a^2}\).
  • Substitution technique: Develops the ability to handle shifted variables systematically.
  • Logarithmic functions: Strengthens understanding of logarithmic terms arising from algebraic integration.
  • Mathematical presentation: Encourages clear algebraic transformations and accurate use of standard formulae.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Tests recognition of integrals involving the standard form \(\sqrt{u^2-a^2}\).
  • Develops fluency in completing the square and selecting suitable substitutions.
  • Reinforces the application of logarithmic expressions in integration.
  • Builds a foundation for more complex integrals involving quadratic radicals.
  • Improves accuracy in handling coefficients and logarithmic arguments.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Complete the square: \(x^2+3x=(x+\frac32)^2-\frac94\).

  2. Recognise the standard form: The transformed expression is of the form \(\sqrt{u^2-a^2}\).

  3. Choose the correct substitution: Take \(u=x+\frac32\), giving \(du=dx\).

  4. Identify the constant: Here, \(a=\frac32\) and \(a^2=\frac94\).

  5. Remember the negative logarithmic term: The formula contains \(-\frac{a^2}{2}\ln|u+\sqrt{u^2-a^2}|\).

  6. Check the real-valued domain: The integrand is real for \(x\leq-3\) or \(x\geq0\).

  7. Verify by differentiation: The derivative of the final expression must reproduce the original integrand.
← Q7
8 / 11  ·  73%
Q9 →
Q9
LONG ANSWER4 marks
Integrate the following function: \[I=\int\sqrt{1+\frac{x^2}{9}}\,dx\]
📘 Concept & Theory
Theory and Concept
›

This problem involves integrating the square root of the sum of two squares. It can be evaluated using the standard formula for the integral of \(\sqrt{u^2+a^2}\).

Standard Integration Formula:

\[\boxed{\int\sqrt{u^2+a^2}\,du=\frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\ln\left|u+\sqrt{u^2+a^2}\right|+C}\]

An equivalent formula, particularly useful when the constant term inside the radical is \(1\), is:

\[\boxed{\int\sqrt{1+u^2}\,du=\frac{u}{2}\sqrt{1+u^2}+\frac12\ln\left|u+\sqrt{1+u^2}\right|+C}\]

In this question, the expression contains \(\frac{x}{3}\). Therefore, substitution is necessary, and the differential factor \(dx=3\,du\) must be included.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rewrite the expression under the square root in terms of \(\frac{x}{3}\).

  2. Substitute \(u=\frac{x}{3}\).

  3. Transform the differential \(dx\) into \(3\,du\).

  4. Apply the standard integration formula for \(\sqrt{1+u^2}\).

  5. Multiply the resulting expression by the factor \(3\).

  6. Substitute \(u=\frac{x}{3}\) back into the result.

  7. Verify the final answer by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  12 steps
  1. Given —
    \[I=\int\sqrt{1+\frac{x^2}{9}}\,dx\]
  2. Since:
    \[\frac{x^2}{9}=\left(\frac{x}{3}\right)^2\]
  3. We can write:
    \[I=\int\sqrt{1+\left(\frac{x}{3}\right)^2}\,dx\]
  4. Apply substitution
  5. Let:
    \[u=\frac{x}{3}\]
  6. Differentiating both sides:
    \[du=\frac13\,dx\]
  7. Therefore:
    \[dx=3\,du\]
  8. Substituting these values into the integral:
    \[ \begin{aligned} I &=\int\sqrt{1+u^2}\,(3\,du)\\ &=3\int\sqrt{1+u^2}\,du \end{aligned} \]
  9. Apply the standard integration formula
  10. We know that:
    \[\int\sqrt{1+u^2}\,du=\frac{u}{2}\sqrt{1+u^2}+\frac12\ln\left|u+\sqrt{1+u^2}\right|+C\]
  11. Therefore:
    \[\begin{aligned}I=3\left[\frac{u}{2}\sqrt{1+u^2}+\frac12\ln\left|u+\sqrt{1+u^2}\right|\right]+C\end{aligned}\]
  12. Multiplying by \(3\):
    \[\begin{aligned}I={}&\frac{3u}{2}\sqrt{1+u^2}\\ &+\frac32\ln\left|u+\sqrt{1+u^2}\right|+C\end{aligned}\]
  13. Substitute \(u=\frac{x}{3}\) back.
  14. Replacing \(u\) with \(\frac{x}{3}\):
    \[\begin{aligned}I={}&\frac{3}{2}\cdot\frac{x}{3}\sqrt{1+\frac{x^2}{9}}\\ &+\frac32\ln\left|\frac{x}{3}+\sqrt{1+\frac{x^2}{9}}\right|+C\end{aligned}\]
  15. Simplifying the first term:
    \[\frac32\cdot\frac{x}{3}=\frac{x}{2}\]
  16. Hence, the final result is:
    \[\boxed{\begin{aligned}\int\sqrt{1+\frac{x^2}{9}}\,dx={}&\frac{x}{2}\sqrt{1+\frac{x^2}{9}}\\ &+\frac32\ln\left|\frac{x}{3}+\sqrt{1+\frac{x^2}{9}}\right|+C\end{aligned}}\]
💡 Answer
Final Answer
›
\[\boxed{\begin{aligned}\int\sqrt{1+\frac{x^2}{9}}\,dx={}&\frac{x}{2}\sqrt{1+\frac{x^2}{9}}\\ &+\frac32\ln\left|\frac{x}{3}+\sqrt{1+\frac{x^2}{9}}\right|+C\end{aligned}}\]
🎯 Exam Significance
Exam Significance
›
  • Standard integration formula: Reinforces the formula for integrating \(\sqrt{1+x^2}\).
  • Substitution technique: Demonstrates the importance of transforming both the variable and its differential.
  • Coefficient management: Highlights how constant factors outside an integral affect every term of the antiderivative.
  • Logarithmic functions: Strengthens understanding of logarithmic terms in algebraic integration.
  • Verification: Demonstrates how differentiation confirms the correctness of an indefinite integral.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Tests recognition of integrals involving the square root of the sum of two squares.
  • Develops accuracy in applying substitutions involving constant multipliers.
  • Reinforces the use of logarithmic expressions in standard integration formulae.
  • Builds a foundation for integrals involving hyperbolic and inverse hyperbolic functions.
  • Highlights the importance of retaining differential factors during substitution.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Recognise the standard form: The expression is \(\sqrt{1+(\frac{x}{3})^2}\).

  2. Choose the substitution: Take \(u=\frac{x}{3}\).

  3. Transform the differential: Since \(du=\frac13dx\), we have \(dx=3du\).

  4. Apply the correct formula: Use the standard formula for \(\sqrt{1+u^2}\).

  5. Retain the multiplier: The factor \(3\) multiplies both terms obtained from integration.

  6. Check the logarithmic coefficient: The correct coefficient is \(\frac32\), not \(\frac12\).

  7. Verify by differentiation: The derivative of the final expression must reproduce the original integrand.
← Q8
9 / 11  ·  82%
Q10 →
Q10
LONG ANSWER4 marks
Evaluate the following integral and select the correct option \[I=\int\sqrt{1+x^2}\,dx\]
  1. \[ \frac{x}{2}\sqrt{1+x^2} +\frac12\log\left|x+\sqrt{1+x^2}\right|+C \]
  2. \[ \frac23(1+x^2)^{3/2}+C \]
  3. \[ \frac23x(1+x^2)^{3/2}+C \]
  4. \[ \frac{x^2}{2}\sqrt{1+x^2} +\frac{x^2}{2}\log\left(x+\sqrt{1+x^2}\right)+C \]
📘 Concept & Theory
Theory and Concept
›

This question tests the application of the standard integration formula involving the square root of the sum of two squares.

The standard formula is:

\[\boxed{\int\sqrt{x^2+a^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln\left|x+\sqrt{x^2+a^2}\right|+C}\]

Comparing the given integral with the standard formula:

\[ a^2=1\quad\Rightarrow\quad a=1 \]

Therefore, the integral can be evaluated directly using this formula.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the standard form \(\sqrt{x^2+a^2}\).

  2. Compare the given expression with the standard integration formula.

  3. Substitute \(a=1\) into the formula.

  4. Compare the resulting expression with the given options.

  5. Verify the correct option by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  3 steps
  1. Given —
    \[I=\int\sqrt{1+x^2}\,dx\]
  2. Recall the standard integration formula
  3. We know that:
    \[\int\sqrt{x^2+a^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln\left|x+\sqrt{x^2+a^2}\right|+C\]
  4. Substitute \(a=1\)
  5. Since \(a^2=1\), we obtain:
    \[ \begin{aligned} I={}& \frac{x}{2}\sqrt{x^2+1}\\ &+\frac12 \ln\left|x+\sqrt{x^2+1}\right|+C \end{aligned} \]
  6. Compare with the given options
  7. The obtained expression is exactly the expression given in Option (A).
    Therefore:
    \[\boxed{\text{Correct Answer: Option (A)}}\]
💡 Answer
Final Answer
›
\[ \boxed{ \int\sqrt{1+x^2}\,dx = \frac{x}{2}\sqrt{1+x^2} + \frac12\ln\left|x+\sqrt{1+x^2}\right|+C } \]
\[ \boxed{\text{Correct Option: (A)}} \]
🎯 Exam Significance
Exam Significance
›
  • Reinforces the standard integration formula for \(\sqrt{x^2+a^2}\).
  • Tests the ability to recognise and apply a formula directly.
  • Develops confidence in verifying antiderivatives through differentiation.
  • Highlights the importance of logarithmic terms in algebraic integration.

For JEE Main, JEE Advanced and Other Competitive Entrance Examinations:

  • Tests rapid recognition of standard integration forms.
  • Develops accuracy in selecting the correct antiderivative from multiple choices.
  • Encourages elimination of incorrect options through differentiation.
  • Strengthens familiarity with logarithmic expressions arising from algebraic integrals.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. The given integral is of the standard form \(\int\sqrt{x^2+a^2}\,dx\).

  2. The value of the constant is \(a=1\).

  3. The standard formula produces both an algebraic term and a logarithmic term.

  4. The coefficient of the logarithmic term is \(\frac{a^2}{2}=\frac12\).

  5. Differentiation confirms the correctness of the resulting antiderivative.
← Q9
10 / 11  ·  91%
Q11 →
Q11
LONG ANSWER4 marks
Evaluate the following integral and identify the correct option \[I=\int\sqrt{x^2-8x+7}\,dx\] \[\begin{aligned}\left( A\right) \dfrac{1}{2}\left( x-4\right) \sqrt{x^{2}-8x+7}+9\log \left| x-4+\sqrt{x^{2}-8x+7}\right| +C\\[5pt] \left( B\right) \dfrac{1}{2}\left( x+4\right) \sqrt{x^{2}-8x+7}+9\log \left| x+4+\sqrt{2^{2}-8x+7}\right| +C\\[5pt] \left( C\right) \dfrac{1}{2}\left( x-4\right) \sqrt{x^{2}-8x+7}+3\sqrt{2}\log \left| x-4\sqrt{x^{2}-8x+7}\right| +C\\[5pt] \left( D\right) \dfrac{1}{2}\left( x-4\right) \sqrt{x^{2}-8x+7}-\dfrac{9}{2}\left| x-4+\sqrt{x^{2}-8x+7}\right| +C\end{aligned}\]
📘 Concept & Theory
Theory and Concept
›

This question involves integration of an algebraic expression containing the square root of a quadratic polynomial. The first step is to complete the square so that the integrand can be expressed in the standard form:

\[ \int\sqrt{u^2-a^2}\,du \]

The standard integration formula is:

\[\boxed{\int\sqrt{u^2-a^2}\,du=\frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\ln\left|u+\sqrt{u^2-a^2}\right|+C}\]

The negative sign before the logarithmic term is important. It distinguishes this formula from the formula for the integral of \(\sqrt{u^2+a^2}\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Complete the square in the quadratic expression.

  2. Introduce a suitable substitution.

  3. Identify the values of \(u\) and \(a\).

  4. Apply the standard integration formula.

  5. Substitute the original variable back.

  6. Verify the result by differentiation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  12 steps
  1. Given —
    \[I=\int\sqrt{x^2-8x+7}\,dx\]
  2. Complete the square.
  3. Consider the quadratic expression:
    \[x^2-8x+7\]
  4. Add and subtract the square of half the coefficient of \(x\):
    \[ \begin{aligned} x^2-8x+7 &=x^2-8x+16-16+7\\ &=(x-4)^2-9 \end{aligned} \]
  5. Therefore:
    \[I=\int\sqrt{(x-4)^2-9}\,dx\]
  6. Apply substitution
  7. Let
    \[u=x-4\]
  8. Differentiating both sides:
    \[du=dx\]
  9. Also,
    \[a^2=9\quad\Rightarrow\quad a=3\]
  10. Substituting these values,
    \[I=\int\sqrt{u^2-3^2}\,du\]
  11. Apply the standard integration formula
  12. Using:
    \[ \begin{aligned} \int\sqrt{u^2-a^2}\,du ={}& \frac{u}{2}\sqrt{u^2-a^2}\\ &-\frac{a^2}{2} \ln\left|u+\sqrt{u^2-a^2}\right|+C \end{aligned} \]
  13. Substitute \(a=3\):
    \[ \begin{aligned} I={}& \frac{u}{2}\sqrt{u^2-9}\\ &-\frac{9}{2} \ln\left|u+\sqrt{u^2-9}\right|+C \end{aligned} \]
  14. Substitute the original variable
  15. Since \(u=x-4\), we obtain:
    \[ \begin{aligned} I={}& \frac{x-4}{2} \sqrt{(x-4)^2-9}\\ &-\frac92 \ln\left|x-4+\sqrt{(x-4)^2-9}\right|+C \end{aligned} \]
  16. Using the identity:
    \[(x-4)^2-9=x^2-8x+7\]
  17. The final result is:
    \[ \boxed{ \begin{aligned} I={}& \frac12(x-4)\sqrt{x^2-8x+7}\\ &-\frac92\ln\left| x-4+\sqrt{x^2-8x+7} \right|+C \end{aligned} } \]
💡 Answer
Final Answer
›

\[ \boxed{ \begin{aligned} \int\sqrt{x^2-8x+7}\,dx ={}& \frac{x-4}{2}\sqrt{x^2-8x+7}\\ &-\frac92\ln\left| x-4+\sqrt{x^2-8x+7} \right|+C \end{aligned} } \]

Correct option: None of the given options as printed. Option (D) becomes correct if its absolute-value term is replaced by the logarithmic term shown above.

🎯 Exam Significance
Exam Significance
›
  • Reinforces completing the square before integration.
  • Tests the application of the standard formula for \(\sqrt{u^2-a^2}\).
  • Highlights the importance of the negative logarithmic term.
  • Develops the ability to verify an indefinite integral through differentiation.

For JEE Main, JEE Advanced and Other Competitive Examinations:

  • Tests recognition of standard algebraic integration forms.
  • Develops accuracy in substitution and coefficient handling.
  • Encourages checking multiple-choice options through differentiation.
  • Strengthens understanding of logarithmic terms in integrals involving quadratic expressions.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Complete the square to simplify quadratic expressions under a square root.

  2. For \(x^2-8x+7\), the correct transformation is \((x-4)^2-9\).

  3. Use the formula for \(\int\sqrt{u^2-a^2}\,du\).

  4. The logarithmic coefficient is \(-\frac{a^2}{2}\), which equals \(-\frac92\) here.

  5. Always check the sign, coefficient and logarithmic function when evaluating multiple-choice answers.
← Q10
11 / 11  ·  100%
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Chapter Complete!

All 11 solutions for Integrals covered.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 12 Maths Integrals Ex 7.7 Solution
NCERT Class 12 Maths Integrals Ex 7.7 Solution — Complete Notes & Solutions · academia-aeternum.com
Explore the step-by-step solution to NCERT Class 12 Mathematics Chapter 7, Integrals, Exercise 7.7, Question 11. This multiple-choice problem focuses on evaluating the integral of the square root of a quadratic expression using the method of completing the square and standard integration formulae. The solution explains how to transform the given quadratic into the form \((x-a)^2-b^2\), apply the standard formula for integrating \(\sqrt{u^2-a^2}\), and obtain the correct antiderivative. Special…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    The integral is ?v(x² - 8x + 7) dx, which can be evaluated by completing the square and applying a standard integration formula.

    Rewrite the expression as x² - 8x + 16 - 16 + 7 = (x - 4)² - 9.

    Use u = x - 4, so that du = dx. The integral becomes ?v(u² - 9) du.

    The formula is ?v(u² - a²) du = (u/2)v(u² - a²) - (a²/2) ln|u + v(u² - a²)| + C.

    Comparing v(u² - 9) with v(u² - a²), we get a² = 9 and therefore a = 3.

    The standard formula for ?v(u² - a²) du contains a negative logarithmic term, -(a²/2) ln|u + v(u² - a²)|. Therefore, the coefficient in this question is -9/2.

    The integral is (x - 4)/2 × v(x² - 8x + 7) - (9/2) ln|x - 4 + v(x² - 8x + 7)| + C.

    None of the options is correct as printed. Option (D) would be correct if its absolute-value expression were replaced by the logarithmic term -(9/2) ln|x - 4 + v(x² - 8x + 7)|.

    Differentiate the obtained antiderivative using the product rule and chain rule. The derivative simplifies to v(x² - 8x + 7), confirming the result.

    It develops skills in completing the square, applying standard integration formulae, handling logarithmic terms and verifying antiderivatives. These techniques are useful in CBSE Board examinations, JEE Main and JEE Advanced.

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