Integrals

• Fundamental substitution technique: This question demonstrates how to recognize a function and its derivative within an integral.
• Standard integral formula: It reinforces the formula for integrating \(1/t\), which is essential for logarithmic integrals.
• Stepwise presentation: Showing the substitution, differential relation, integration and back-substitution makes the method clear and supports stepwise marking.
• Verification skill: Differentiating the final result is a reliable way to check an indefinite integral.

Significance for Competitive Entrance Examinations

• Pattern recognition: Competitive examinations frequently test whether students can identify an integrand of the form \(f'(x)/f(x)\).
• Time-efficient calculation: Recognizing the logarithmic pattern avoids unnecessary algebraic manipulation.
• Foundation for advanced problems: The same principle is useful in more complicated integrals involving rational functions, trigonometric expressions and composite functions.
• Common pattern to remember: Whenever the numerator is the derivative of the denominator, check whether logarithmic integration applies.

• Application of substitution: This problem tests the ability to identify an inner function and its derivative within an integral.
• Use of the power rule: It reinforces the integration of polynomial powers after simplifying a logarithmic expression.
• Stepwise presentation: Clearly showing the substitution, differential relation, integration and back-substitution makes the reasoning easy to follow and supports stepwise marking.
• Chain rule connection: The verification demonstrates how differentiation reverses the integration process.
• Domain awareness: Recognizing that \(\log x\) is defined for real \(x>0\) helps maintain mathematical accuracy.

• Quick pattern recognition: Integrals of the form \(\int \frac{(\log x)^n}{x}\,dx\) can be handled efficiently using the substitution \(t=\log x\).
• Generalisation: The same method gives a useful formula for higher powers of logarithms.
• Reduced calculation: Substitution converts a logarithmic integral into a straightforward power integral, avoiding unnecessary integration techniques.
• Conceptual preparation: The method develops skills needed for more complicated integrals involving composite functions and their derivatives.

• Factorisation skills: The question tests the ability to simplify an algebraic denominator before applying an integration formula.
• Substitution method: It reinforces the recognition of a function and its derivative within the same integral.
• Standard logarithmic integral: The solution uses \(\int \frac{1}{t}\,dt=\ln|t|+C\), a fundamental formula in the Integrals chapter.
• Mathematical precision: Retaining the absolute value in the logarithmic answer demonstrates correct handling of the domain.
• Stepwise working: Showing each transformation clearly helps communicate the reasoning and supports stepwise marking.

• Pattern recognition: The integral follows the general structure \(\int \frac{f'(x)}{f(x)}\,dx\), which leads to a logarithmic result.
• Efficient calculation: Factoring the denominator and choosing the correct substitution reduces the problem to one standard integration formula.
• Conceptual understanding: The problem connects algebraic simplification, logarithmic differentiation and integration by substitution.
• Domain awareness: Identifying where the denominator vanishes helps avoid invalid substitutions and incorrect conclusions.
• Transfer to advanced problems: The same technique applies to rational expressions and composite functions whose derivatives appear in the numerator or differential factor.

• Integration by substitution: The problem tests the ability to identify a composite function and select its inner function as the substitution.
• Trigonometric integration: It reinforces the standard formula for integrating \(\sin t\).
• Sign accuracy: The negative sign arising from the derivative of \(\cos x\) is an important detail in stepwise solutions.
• Chain rule connection: The solution demonstrates how integration reverses the differentiation of a composite function.
• Presentation of working: Showing the substitution, differential relation, integration and back-substitution makes the solution logically complete.

• Rapid pattern recognition: Integrals involving a trigonometric function of another trigonometric function often become simpler when the inner function is substituted.
• Chain rule in reverse: Recognising the derivative of the inner function helps solve composite-function integrals efficiently.
• Sign management: Careful handling of negative signs is essential for avoiding errors in timed examinations.
• Generalisation: The same technique applies to many integrals involving expressions such as \(\sin(f(x))f'(x)\) or \(\cos(f(x))f'(x)\).

These formulas follow from the chain rule and are useful for recognising composite-function integrals quickly.

• Substitution technique: The problem tests the ability to simplify a trigonometric integral by introducing a suitable variable.
• Chain rule application: Differentiating \(ax+b\) introduces the constant factor \(a\), which must be included in the substitution.
• Sign accuracy: The negative sign in the derivative of cosine must be handled correctly throughout the calculation.
• Use of standard formulas: The problem reinforces the power rule of integration and the differentiation of composite trigonometric functions.
• Stepwise presentation: Showing the differential relation explicitly helps avoid missing factors and makes the solution mathematically complete.

• Recognition of composite functions: Expressions involving \(ax+b\) are common in integration problems and require careful use of the chain rule.
• Constant-factor awareness: Missing the factor \(\frac{1}{a}\) is a common error when integrating functions of linear expressions.
• Alternative solution strategies: The identity \(\sin\theta\cos\theta=\frac{1}{2}\sin 2\theta\) provides another approach and helps students compare methods.
• Verification skill: Differentiating an antiderivative is an efficient way to identify sign errors and missing constant factors.

An equivalent useful formula, for \(a\neq 0\), is

The two forms are equivalent because \(\sin^2\theta+\cos^2\theta=1\); their difference is only a constant, which is absorbed into \(C\).

• Substitution method: The problem tests the ability to simplify an integral containing a linear expression under a radical.
• Fractional powers: It reinforces the conversion of radicals into fractional exponents before applying the power rule.
• Handling constants: The factor \(\frac{1}{a}\) arises from the differential substitution and must be retained throughout the solution.
• Stepwise presentation: Clearly showing the substitution, differential relation, integration and back-substitution makes the derivation easy to follow.
• Verification: Differentiating the result confirms the correctness of the antiderivative.

• Pattern recognition: Integrals containing powers of \(ax+b\) can often be evaluated directly using substitution.
• Efficient calculation: Replacing the inner linear expression with \(t\) converts the problem into a standard power integral.
• Chain rule connection: Understanding why the coefficient \(\frac{1}{a}\) appears helps prevent common errors.
• Generalisation: The same method works for other powers of linear expressions, not just square roots.

• Integration by substitution: The problem tests the ability to select a substitution that simplifies a radical expression.
• Algebraic rearrangement: Expressing \(x\) as \(t^2-2\) is an important step in transforming the entire integrand.
• Polynomial integration: The transformed integral reinforces the power rule and the linearity of integration.
• Accurate expansion: Expanding \((t^2-2)t^2\) and retaining the coefficients correctly prevents common calculation errors.
• Stepwise presentation: Clearly showing the substitution, differential, expansion, integration and back-substitution makes the solution complete and easy to follow.

• Choosing an effective substitution: Replacing a radical with a new variable is a useful strategy for simplifying algebraic integrals.
• Transformation of expressions: The question develops the ability to convert a mixed algebraic-radical expression into a polynomial.
• Calculation efficiency: Once the substitution is chosen correctly, the remaining integration involves only standard power rules.
• Error prevention: Tracking all three substitutions, \(x=t^2-2\), \(\sqrt{x+2}=t\) and \(dx=2t\,dt\), prevents missing factors.
• Transferable technique: The same method is useful for integrals involving products of algebraic expressions and radicals such as \(\sqrt{ax+b}\).

• Substitution technique: The problem demonstrates how to simplify an integral containing a square root of a quadratic expression.
• Chain rule application: Differentiating \(t^2=1+2x^2\) requires careful application of the chain rule.
• Differential manipulation: Expressing \(x\,dx\) in terms of \(t\,dt\) is the crucial step in simplifying the integral.
• Power rule: The transformed integral can be evaluated using a standard integration formula.
• Stepwise presentation: Showing every substitution and algebraic transformation makes the solution clear and complete.

• Pattern recognition: Expressions of the form \(x\sqrt{a+bx^2}\) often suggest a substitution involving the expression under the radical.
• Efficient integration: The substitution reduces the original integral to a single power integral, avoiding unnecessary expansion.
• Coefficient accuracy: Correctly deriving \(x\,dx=\frac{t}{2}\,dt\) prevents errors in the final numerical factor.
• Generalisation: The method can be extended to integrals involving powers of quadratic expressions such as \(x(a+bx^2)^n\).

A useful general formula is

where \(b\neq 0\) and \(n\neq -1\), on intervals where the power is defined. The given problem corresponds to \(a=1\), \(b=2\) and \(n=\frac{1}{2}\).

• Recognising a derivative pattern: The expression \(4x+2\) is twice the derivative of \(x^2+x+1\), making substitution the natural method.
• Chain rule application: Differentiating \(t^2=x^2+x+1\) requires the correct factor \(2t\).
• Coefficient management: Rewriting \(4x+2\) as \(2(2x+1)\) makes the factor of \(4\) in the transformed integral clear.
• Fractional exponents: The problem tests the correct conversion of \(t^3\) into \((x^2+x+1)^{3/2}\).
• Stepwise presentation: Showing each transformation helps prevent missing coefficients and incorrect exponents.

• Pattern recognition: Integrals of the form \(f'(x)\sqrt{f(x)}\) can often be evaluated directly by substitution.
• Calculation efficiency: The correct substitution avoids expanding or manipulating the quadratic expression unnecessarily.
• Error prevention: Keeping track of the constant multiplier and the exponent is essential for accuracy in timed examinations.
• Generalisation: The same method applies to integrals involving other powers of quadratic polynomials.

• Substitution involving radicals: The question demonstrates how to eliminate a square root by introducing a new variable.
• Factorisation: Expressing \(t^2-t\) as \(t(t-1)\) is essential for simplifying the integral.
• Logarithmic integration: The problem reinforces the standard formula for integrating the reciprocal of a linear expression.
• Absolute value: Writing \(\ln|t-1|\) correctly handles the logarithm on both sides of \(t=1\).
• Stepwise presentation: Showing the substitution, cancellation and logarithmic integration clearly helps avoid algebraic and sign errors.

• Radical substitution: Expressions containing both \(x\) and \(\sqrt{x}\) often become rational functions after substituting \(t=\sqrt{x}\).
• Efficient simplification: Factoring the denominator immediately reveals the cancellation that reduces the integral to a standard form.
• Recognition of logarithmic patterns: The transformed integrand has the form \(\frac{f'(t)}{f(t)}\), which integrates to a logarithm.
• Domain awareness: Identifying points where the original denominator vanishes prevents invalid manipulations and incorrect domain assumptions.
• Transferable method: Similar substitutions can simplify integrals involving rational expressions in \(\sqrt{x}\).

A useful general result is

This formula applies on intervals where \(f\) is differentiable and nonzero.

• Substitution involving radicals: The question develops the skill of eliminating a square root through an appropriate variable substitution.
• Correct differential handling: The relationship \(dx=2t\,dt\) must be used consistently with the denominator.
• Algebraic simplification: Cancelling the common factor before integrating prevents unnecessary logarithmic terms.
• Power rule application: The transformed integral reinforces the integration of polynomial expressions.
• Error detection: Differentiation verifies the result and helps identify errors caused by incorrect cancellation.

• Strategic substitution: Choosing \(t=\sqrt{x+4}\) transforms the radical into a simple variable.
• Efficient simplification: Recognising that the factor introduced by \(dx\) cancels the denominator avoids unnecessary algebra.
• Accuracy under time pressure: Tracking the numerator, denominator and differential separately reduces common integration errors.
• Method selection: The problem illustrates why substitution should be followed by a careful simplification before selecting an integration formula.
• Verification skills: Differentiation is a quick and reliable method for checking the correctness of an antiderivative.

• Substitution technique: The question demonstrates how to simplify a fractional power of a polynomial by introducing a new variable.
• Algebraic decomposition: Expressing \(x^5\,dx\) as \(x^3(x^2\,dx)\) is the key step that makes the substitution effective.
• Fractional-power integration: The problem reinforces the power rule for rational exponents.
• Sign and coefficient accuracy: Students must expand the expression correctly and apply the factor \(\frac13\) to both terms.
• Stepwise presentation: Clearly showing each substitution and integration step helps demonstrate the complete method.

• Pattern recognition: Expressions containing \(x^3-1\) and a high power of \(x\) often suggest substituting \(t=x^3-1\).
• Efficient algebraic manipulation: Separating \(x^5\,dx\) into \(x^3(x^2\,dx)\) avoids solving for \(x\) explicitly.
• Fractional exponent fluency: Converting between fractional exponents and power-rule expressions is useful in JEE and other entrance examinations.
• Error prevention: Careful handling of signs, coefficients and exponents is essential in multi-step integration problems.
• Verification skill: Differentiating the answer provides a reliable way to check a result and identify algebraic mistakes.

• Substitution method: The question tests the ability to identify an inner polynomial and its derivative in the integrand.
• Negative powers: It reinforces the power rule for integrating expressions with negative exponents.
• Constant factors: The factor \(\frac19\) comes from differentiating \(2+3x^3\) and must be retained throughout the calculation.
• Correct back-substitution: Replacing \(t\) with the original expression prevents mistakes involving the power of \(x\).
• Stepwise presentation: Showing the differential relation and the integration formula makes the method clear and complete.

• Pattern recognition: Integrals of the form \(\int x^2(a+bx^3)^n\,dx\) naturally suggest substitution of the cubic expression.
• Efficient calculation: The substitution converts a rational function into a simple power integral.
• Accuracy with exponents: Correctly integrating \(t^{-3}\) gives a term proportional to \(t^{-2}\), not a logarithm.
• Generalisation: The method applies to a wide range of integrals involving a polynomial function and its derivative.

A useful general result is

where \(b\neq 0\) and \(n\neq -1\), on intervals where the expression is defined.

• Substitution technique: The question tests recognition of a function and its derivative within an integral.
• Logarithmic differentiation: Students must know that \(\frac{d}{dx}(\log x)=\frac1x\) for \(x>0\).
• Power rule: It reinforces integration of a general power \(t^{-m}\).
• Parameter restriction: The condition \(m\neq1\) is essential because the power-rule denominator \(1-m\) must not be zero.
• Domain awareness: The original integrand is undefined at \(x=1\), so the integration constant may be different on the intervals \(0 < x < 1\) and \(x>1\).

• Pattern recognition: An integrand containing \(\frac1x\) multiplied by a function of \(\log x\) suggests the substitution \(t=\log x\).
• Generalisation: The result works for any real parameter \(m\neq1\), wherever the real-valued integrand is defined.
• Special-case awareness: If \(m=1\), the power rule cannot be used in this form. Instead, the integral becomes \(\int\frac{dt}{t}\).
• Efficient evaluation: The substitution reduces a logarithmic integral to a standard integration formula, saving time in objective examinations.

Special case when \(m=1\):

• Substitution method: The question tests whether students can recognise the relationship between the numerator and the derivative of the denominator.
• Logarithmic integration: It applies the standard formula \(\int\frac1t\,dt=\log|t|+C\).
• Sign accuracy: Differentiating \(9-4x^2\) produces \(-8x\), so the final answer must contain a negative coefficient.
• Absolute value: Students should retain the absolute value inside the logarithm.
• Presentation: Clearly showing the substitution and differential relation makes the solution easy to follow and supports stepwise marking.

• Quick recognition: For integrals of the form \(\int\frac{x}{a-bx^2}\,dx\), substitution of the denominator is often the fastest approach.
• General formula: For constants \(a,b\) with \(b\neq0\),

• Common-error prevention: The negative sign arises from the derivative of \(a-bx^2\); omitting it changes the derivative of the result.
• Verification: Differentiation provides a quick way to confirm the sign and coefficient, especially in multiple-choice questions.

• Substitution method: The question tests the use of substitution to simplify an exponential integral.
• Chain rule connection: Students must account for the derivative of the exponent when integrating.
• Constant factor: The factor \(\frac12\) is essential because the derivative of \(2x+3\) is \(2\).
• Standard formula: The problem reinforces the identity \(\int e^x\,dx=e^x+C\).
• Verification: Differentiating the final answer confirms that the coefficient is correct.

• Rapid evaluation: Exponential functions with linear exponents can be integrated directly using a standard formula.
• Generalisation: For \(a\neq0\), the formula is

• Common-error prevention: Omitting the factor \(\frac1a\) leads to an incorrect derivative.
• Chain rule awareness: The same principle applies to more complicated exponential integrals when the derivative of the exponent is present or can be accounted for.

• Substitution method: The problem tests recognition of the relationship between \(x\) and the exponent \(x^2\).
• Exponent laws: Rewriting \(\frac1{e^{x^2}}\) as \(e^{-x^2}\) simplifies the calculation.
• Sign accuracy: Integrating \(e^{-t}\) introduces a negative sign, which must be retained.
• Chain rule: Differentiation of the final answer confirms both the coefficient and the sign.
• Stepwise presentation: Showing the substitution, integration formula and back-substitution makes the reasoning clear.

• Quick recognition: Integrals of the form \(\int xe^{-x^2}\,dx\) suggest substituting \(t=x^2\).
• General formula: For a nonzero constant \(a\),

• Common-error prevention: A positive answer \(\frac12e^{-x^2}\) differentiates to \(-xe^{-x^2}\), which has the wrong sign.
• Efficient verification: Differentiating the candidate antiderivative is a reliable way to catch sign errors in objective questions.

• Inverse trigonometric derivatives: The question tests knowledge of the derivative of \(\tan^{-1}x\).
• Substitution technique: Students must identify that the derivative of the exponent is already present in the integrand.
• Exponential integration: It applies the standard formula \(\int e^t\,dt=e^t+C\).
• Chain rule: Differentiating the final answer confirms that the inverse tangent derivative has been handled correctly.
• Clear presentation: Showing the substitution and differential relation makes the solution concise and logically complete.

• Pattern recognition: An expression of the form \(e^{f(x)}f'(x)\) can be integrated directly as \(e^{f(x)}+C\).
• General formula:

• Efficient problem-solving: Recognising the derivative of the exponent avoids unnecessary algebraic manipulation.
• Cross-topic application: The problem combines inverse trigonometric differentiation with exponential integration, a useful pattern for entrance examinations.

• Algebraic simplification: Dividing by \(e^x\) converts the expression into a convenient form for substitution.
• Exponential derivatives: Students must remember that the derivative of \(e^{-x}\) is \(-e^{-x}\).
• Logarithmic integration: The method applies the formula \(\int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C\).
• Sign accuracy: Correctly differentiating the substitution is essential for matching the numerator.
• Verification: Differentiation confirms that the simplified logarithmic expression is correct.

• Pattern recognition: Look for an expression whose derivative appears in the numerator.
• Efficient substitution: The substitution \(t=e^x+e^{-x}\) reduces the integral directly to a logarithm.
• General formula:

• Common-error prevention: The negative sign in \(\frac{d}{dx}(e^{-x})=-e^{-x}\) is essential.
• Alternative approach: Recognising the logarithmic derivative pattern can save time in JEE and other entrance examinations.

• Substitution method: The question tests whether students can identify a denominator whose derivative is proportional to the numerator.
• Exponential differentiation: It reinforces the chain rule for \(e^{2x}\) and \(e^{-2x}\).
• Logarithmic integration: The problem uses the formula \(\int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C\).
• Constant factor: The factor \(\frac12\) arises because the derivative of the denominator is twice the numerator.
• Verification: Differentiation confirms both the coefficient and the signs in the final answer.

• Pattern recognition: When the numerator resembles the derivative of the denominator, try the denominator as a substitution.
• General formula: If \(f(x)\neq0\) and the numerator is proportional to \(f'(x)\), then

• Efficient evaluation: Recognising this pattern avoids unnecessary algebraic transformations.
• Common-error prevention: Remember that differentiating \(e^{-2x}\) produces a negative factor, \(-2e^{-2x}\).

• Substitution method: The question tests the ability to simplify a trigonometric integral with a linear argument.
• Trigonometric identity: Students must remember \(\tan^2 t=\sec^2 t-1\).
• Integration formulas: It applies \(\int\sec^2 t\,dt=\tan t+C\).
• Constant factors: The factor \(\frac12\) arises from differentiating \(2x-3\).
• Constants of integration: The constant \(\frac32\) can be incorporated into \(C\), simplifying the final answer.

• Identity recognition: For an even power of tangent, the identity \(1+\tan^2 t=\sec^2 t\) often provides a direct route to integration.
• General formula: For \(a\neq0\),

• Common-error prevention: The substitution factor must multiply both terms obtained after applying the identity.
• Fast verification: Differentiating the final expression quickly confirms that the coefficient and the linear term are correct.

• Substitution method: The question tests the use of substitution for a trigonometric function with a linear argument.
• Standard formula: It applies \(\int\sec^2t\,dt=\tan t+C\).
• Negative coefficient: The factor \(-\frac14\) arises from differentiating \(7-4x\) and must not be omitted.
• Chain rule: Differentiating the final answer confirms the correct sign and coefficient.
• Stepwise presentation: Clearly showing the substitution and differential relation supports a complete solution.

Significance for Competitive Entrance Examinations

• Rapid evaluation: Integrals of the form \(\int\sec^2(ax+b)\,dx\) can be solved directly by substitution.
• General formula: For \(a\neq0\),

• Sign-error prevention: A negative coefficient in the argument produces a negative factor in the antiderivative.
• Verification: Differentiation is an efficient way to confirm the result in objective examinations.

• Inverse trigonometric derivatives: The question tests knowledge of the derivative of \(\sin^{-1}x\).
• Substitution method: Students must identify that the derivative of the numerator's inverse trigonometric function appears in the integrand.
• Power rule: The transformed integral is a simple polynomial integral.
• Chain rule: Verification demonstrates how the derivative of a squared function is calculated.
• Notation: Once integration is complete, write \(\frac{t^2}{2}+C\), without an additional \(dt\).

Significance for Competitive Entrance Examinations

• Pattern recognition: An integrand of the form \(f(x)f'(x)\) can often be integrated by treating \(f(x)\) as a substitution.
• General formula:

Here, \(f(x)=\sin^{-1}x\) and \(f'(x)=\frac{1}{\sqrt{1-x^2}}\).

• Efficient evaluation: Recognising the derivative pattern avoids unnecessary algebraic transformations.
• Domain awareness: The integrand is real and defined for \(-1

• Substitution method: The question tests recognition of a denominator whose derivative is proportional to the numerator.
• Trigonometric differentiation: It reinforces the derivatives of \(\sin x\) and \(\cos x\).
• Logarithmic integration: It applies \(\int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C\).
• Constant factor: The factor \(\frac12\) must be retained because the derivative of the denominator is twice the numerator.
• Absolute value: The logarithm must contain the absolute value of the denominator.

Significance for Competitive Entrance Examinations

• Pattern recognition: For an integral of the form \(\int\frac{f'(x)}{f(x)}\,dx\), the result is logarithmic.
• General formula:

• Efficient evaluation: Recognising the relationship between numerator and denominator avoids lengthy trigonometric manipulations.
• Error prevention: Differentiation provides a quick check of the coefficient and the signs.

• Substitution method: The problem tests recognition of a function and its derivative within a trigonometric integral.
• Trigonometric derivatives: It reinforces \(\frac{d}{dx}(\tan x)=\sec^2x\).
• Power rule: The transformed integral requires integrating \(t^{-2}\).
• Sign handling: The negative sign in \(dt=-\sec^2x\,dx\) cancels with the negative sign arising from the power-rule denominator.
• Verification: Differentiating the final answer confirms both the sign and the exponent.

Significance for Competitive Entrance Examinations

• Pattern recognition: When an integrand contains \(\sec^2x\) and a power of \(1-\tan x\), substituting \(t=1-\tan x\) is efficient.
• General formula: For \(n\neq-1\),

For this question, \(a=1\) and \(n=2\), giving \(\frac{1}{1-\tan x}+C\).

• Common-error prevention: Keep track of the negative sign introduced by differentiating \(1-\tan x\).
• Efficient verification: The chain rule quickly confirms the final expression.

• Substitution technique: The question tests the ability to simplify an integral containing a square-root expression.
• Derivative of a square root: Students must remember that \(\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\).
• Trigonometric integration: It applies the standard formula \(\int\cos t\,dt=\sin t+C\).
• Constant factor: The factor \(2\) comes from rearranging the differential relation and must not be omitted.
• Sign accuracy: The positive derivative of \(\sqrt{x}\) ensures the final coefficient is positive.

Significance for Competitive Entrance Examinations

• Pattern recognition: For integrals containing a function of \(\sqrt{x}\) multiplied by \(\frac1{\sqrt{x}}\), substituting \(t=\sqrt{x}\) is often effective.
• General formula: For a differentiable function \(f\),

The formula is understood with the substitution applied to the right-hand integral.

• Error prevention: Check the sign when differentiating the inner function before transforming the integral.
• Fast verification: Differentiating the final answer quickly confirms both the coefficient and the sign.

• Substitution technique: The problem tests recognition of the derivative of \(\sin 2x\) within the integrand.
• Chain rule: The factor \(2\) appears when differentiating \(\sin 2x\), so the substitution contributes a factor of \(\frac12\).
• Fractional powers: It reinforces the integration rule for \(t^{1/2}\).
• Back-substitution: The original inner expression \(\sin 2x\) must be restored accurately.
• Verification: Differentiation checks both the coefficient \(\frac13\) and the argument \(2x\).

Significance for Competitive Entrance Examinations

• Pattern recognition: An integrand of the form \([f(x)]^n f'(x)\) suggests a direct substitution.
• General formula: For \(n\neq-1\),

• Efficient evaluation: This pattern avoids unnecessary trigonometric identities.
• Common-error prevention: Carefully track the derivative of the inner function and substitute the entire expression back into the answer.

• Substitution: Tests the ability to identify an expression whose derivative appears elsewhere in the integrand.
• Fractional powers: Reinforces the power rule for negative fractional exponents.
• Algebraic simplification: Students must correctly simplify division by \(\frac12\) to multiplication by \(2\).
• Presentation: Clearly showing the substitution, differential relation, integration, and back-substitution helps communicate each step of the solution.

Significance for Competitive Entrance Examinations

• Recognising standard patterns: The integral has the form \(\int \frac{f'(x)}{\sqrt{f(x)}}\,dx\).
• General result: For \(f(x)>0\),

• Speed and accuracy: Recognising this pattern can reduce a multistep integral to a direct substitution.
• Error prevention: The factor \(2\) arises from the power rule and must not be omitted.

• Substitution method: Tests recognition of a function whose derivative appears as another factor in the integrand.
• Logarithmic differentiation: Reinforces the derivative formula \(\frac{d}{dx}\log u=\frac{u'}{u}\).
• Trigonometric identities: Uses \(\frac{\cos x}{\sin x}=\cot x\).
• Stepwise presentation: Clearly showing the substitution, differential relation, integration, and back-substitution makes the reasoning easy to follow.

• Recognising a standard pattern: The integral has the form \(\int f'(x)f(x)\,dx\), where \(f(x)=\log(\sin x)\).
• General formula:

• Efficient evaluation: Substitution reduces the integral directly to a simple polynomial integral.
• Error prevention: Remember that the derivative of \(\log(\sin x)\) is \(\cot x\), not \(\tan x\).

• Substitution technique: Demonstrates how to simplify a trigonometric integral by substituting its denominator.
• Sign management: The negative sign arises from differentiating \(\cos x\), since \(\frac{d}{dx}(\cos x)=-\sin x\).
• Logarithmic integration: Reinforces the standard result \(\int\frac{1}{t}\,dt=\ln|t|+C\).
• Domain awareness: The original denominator must be nonzero, so \(x\neq(2k+1)\pi\), where \(k\in\mathbb Z\).

• Recognising a standard pattern: Integrals of the form \(\int\frac{f'(x)}{f(x)}\,dx\) lead to logarithmic functions.
• General formula:

• Quick evaluation: Recognising the derivative of the denominator avoids unnecessary trigonometric transformations.
• Common mistake prevention: Keep the negative sign from \(dt=-\sin x\,dx\) throughout the calculation.

• Substitution technique: Demonstrates how a trigonometric integral can be converted into an algebraic power integral.
• Power rule: Tests the integration of negative powers, particularly \(t^{-2}\).
• Sign accuracy: Students must account for both the negative sign from \(dt=-\sin x\,dx\) and the negative denominator in the power-rule formula.
• Verification: Differentiation confirms the result and helps identify sign errors.

Significance for Competitive Entrance Examinations

• Pattern recognition: The integral has the form \(\int\frac{f'(x)}{[f(x)]^2}\,dx\), up to a constant sign.
• General formula:

In this problem, \(f(x)=1+\cos x\), so \(f'(x)=-\sin x\). Thus, the negative sign in the general formula cancels the negative sign in \(f'(x)\), yielding the positive final result.

• Efficient evaluation: Recognising the reciprocal-square pattern avoids unnecessary trigonometric identities.
• Common mistake prevention: Keep track of signs during both substitution and integration.

• Trigonometric simplification: Tests the ability to express cotangent in terms of sine and cosine.
• Algebraic decomposition: The numerator is split to produce a constant term and the derivative of the denominator, up to a sign.
• Logarithmic integration: Reinforces the formula \(\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C\).
• Sign accuracy: The substitution produces a negative sign because \(dt=(\cos x-\sin x)\,dx\).

Significance for Competitive Entrance Examinations

• Recognising the derivative pattern: Look for an expression whose derivative appears in the numerator after algebraic decomposition.
• Efficient integration: Splitting the integrand avoids more complicated trigonometric substitutions.
• Common-error prevention: Distinguish between \(\int\frac{1}{t}\,dt=\ln|t|+C\) and \(\int\ln t\,dt=t\ln t-t+C\).
• Domain awareness: The logarithmic term requires \(\sin x+\cos x\neq0\), consistent with the domain of the original integrand.

• Trigonometric simplification: Tests the conversion of tangent into sine and cosine.
• Algebraic decomposition: Splitting the numerator produces a constant term and a term involving the derivative of the denominator.
• Logarithmic integration: Reinforces the formula \(\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C\).
• Sign accuracy: Differentiating \(\cos x-\sin x\) gives \(-\sin x-\cos x\), which determines the sign of the logarithmic term.

Significance for Competitive Entrance Examinations

• Pattern recognition: Look for a way to rewrite the numerator using the denominator and its derivative.
• Efficient evaluation: This decomposition avoids more complicated trigonometric substitutions.
• Common-error prevention: Distinguish between \(\int\frac{1}{t}\,dt=\ln|t|+C\) and \(\int\ln t\,dt=t\ln t-t+C\).
• Domain awareness: Check where the original denominator is nonzero before interpreting the logarithmic result.

• Trigonometric identities: Tests the conversion between tangent, sine, cosine, and secant.
• Substitution method: Reinforces the use of \(t=\tan x\) when \(\sec^2 x\,dx\) appears in the integral.
• Fractional powers: Requires correct integration of \(t^{-1/2}\).
• Stepwise reasoning: Rewriting the integrand before substitution makes the solution clearer and reduces algebraic errors.

Significance for Competitive Entrance Examinations

• Recognising the derivative pattern: The derivative of \(\tan x\) is \(\sec^2 x\), which is the key to the substitution.
• Useful identity:

• Efficient evaluation: This identity converts the integral directly into a power integral.
• Common-error prevention: Do not introduce an extra cosine factor or alter the argument of the square root during substitution.

• Substitution method: Tests recognition of an inner function whose derivative appears as a factor in the integrand.
• Logarithmic differentiation: Reinforces \(\frac{d}{dx}(\log x)=\frac1x\).
• Power rule: Applies the standard integration formula to a cubic expression after substitution.
• Stepwise presentation: Showing the substitution, differential relation, integration, and back-substitution makes the solution easy to follow.

Significance for Competitive Entrance Examinations

• Recognising a standard pattern: The integral has the form \(\int[f(x)]^n f'(x)\,dx\).
• General formula:

• Efficient evaluation: A suitable substitution converts a logarithmic integral into a simple polynomial integral.
• Common-error prevention: Keep the factor \(\frac1x\) associated with \(dx\) when converting to \(dt\).

• Composite-function substitution: Tests recognition of an expression and its derivative within the same integrand.
• Logarithmic differentiation: Uses \(\frac{d}{dx}(\log x)=\frac1x\).
• Power rule: Reduces the problem to a standard polynomial integral.
• Stepwise presentation: Writing \(dt=\frac{x+1}{x}\,dx\) explicitly shows why the substitution works.

Significance for Competitive Entrance Examinations

• Pattern recognition: The integral has the form \(\int[f(x)]^n f'(x)\,dx\).
• General formula:

• Efficient evaluation: Recognising the derivative factor avoids expanding \((x+\log x)^2\).
• Common-error prevention: Keep the factor \(\frac{x+1}{x}\) together with \(dx\) when substituting for \(dt\).

• Inverse trigonometric derivatives: Tests the formula for differentiating \(\tan^{-1}u\).
• Chain rule: Requires differentiating the composite function \(\tan^{-1}(x^4)\).
• Trigonometric integration: Uses the standard result \(\int\sin t\,dt=-\cos t+C\).
• Sign accuracy: A missing negative sign changes the derivative and makes the answer incorrect.

Significance for Competitive Entrance Examinations

• Pattern recognition: Recognise the derivative of \(\tan^{-1}(x^4)\) within the integrand.
• General formula:

• Efficient evaluation: A suitable substitution converts a composite trigonometric integral into a standard integral.
• Equivalent forms: The inverse-trigonometric expression can be simplified using a right-triangle identity or the identity for \(\cos(\tan^{-1}u)\).

• Tests the logarithmic integration formula.
• Combines differentiation of a polynomial and an exponential function.
• Checks recognition of the derivative of the denominator in the numerator.

Significance for Competitive Entrance Examinations

• Quick recognition: If the numerator is the derivative of the denominator, use \(\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C\).
• Common-error prevention: Remember that \(\frac{d}{dx}(a^x)=a^x\ln a\), so \(\frac{d}{dx}(10^x)=10^x\ln 10\).
• Time efficiency: Recognising the pattern avoids unnecessary expansion or substitution steps.