JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Indefinite Integration Using Integration by Parts and Chain Rule

Q: Why choose e^(sin^2 x)sin 2x as the second function rather than expanding sin 2x = 2sin x cos x?

Because the derivative of sin^2 x is 2sin x cos x = sin 2x, making integral of e^(sin^2 x)sin 2x dx = e^(sin^2 x) directly integrable.

If I(x)=integral of e^(sin^2 x)(cos xsin 2x-sin x)dx and I(0)=1, then I(π/3) is equal to

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A-1/2e^(3/4)Be^(3/4)C1/2e^(3/4)D-e^(3/4)

Step-by-step solutionView

Correct answer

By recognizing the integrand as the derivative of $\cos x \cdot e^{\sin^{2} x}$ , we find $I (x) = \cos x \cdot e^{\sin^{2} x}$ , which gives $I (\frac{π}{3}) = \frac{1}{2} e^{\frac{3}{4}}$ .

Option analysis

Why each option works or fails

A · -1/2e^(3/4)

Carrying an extra negative sign through the product rule differentiation or integration steps leads to $- \frac{1}{2} e^{3 / 4}$ . Check the derivative of $\cos x \cdot e^{\sin^{2} x}$ : $\frac{d}{d x} (\cos x) = - \sin x$ , so the negative sign is already correctly accounted for in the integrand.

B · e^(3/4)

Evaluating $\cos (\frac{π}{3})$ incorrectly as $1$ instead of $\frac{1}{2}$ . Recall standard trigonometric values: $\cos (\frac{π}{3}) = \frac{1}{2}$ and $\sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ .

C · 1/2e^(3/4)

This is the correct value obtained from $I (x) = \cos x \cdot e^{\sin^{2} x}$ evaluated at $x = \frac{π}{3}$ . Correctly integrated and evaluated.

D · -e^(3/4)

Combining both a sign error and an incorrect trigonometric evaluation of $\cos (\frac{π}{3}) = - 1$ . Carefully compute both the antiderivative via product rule inspection and the standard trigonometric values at $\frac{π}{3}$ .

Reviewed route

Solution

StepWorking

01Given

$I (x) = \int e^{\sin^{2} x} (\cos x \sin 2 x - \sin x) d x$ with initial condition $I (0) = 1$ .

02Goal

Evaluate $I (\frac{π}{3})$ .

03Approach

Split the integrand into $I_{1} = \int e^{\sin^{2} x} \sin 2 x \cos x d x$ and $I_{2} = - \int e^{\sin^{2} x} \sin x d x$ . Apply integration by parts on $I_{1}$ taking $\cos x$ as the first function and $e^{\sin^{2} x} \sin 2 x$ as the second function because $\frac{d}{d x} (e^{\sin^{2} x}) = e^{\sin^{2} x} \sin 2 x$ .

04Execute

Integrating $I_{1}$ by parts: $\int \cos x \cdot (e^{\sin^{2} x} \sin 2 x) d x = \cos x \cdot e^{\sin^{2} x} - \int (- \sin x) e^{\sin^{2} x} d x = e^{\sin^{2} x} \cos x + \int e^{\sin^{2} x} \sin x d x$ Combining with $I_{2}$ : $I (x) = e^{\sin^{2} x} \cos x + \int e^{\sin^{2} x} \sin x d x - \int e^{\sin^{2} x} \sin x d x = e^{\sin^{2} x} \cos x + C$

05Execute

Using $I (0) = 1$ : $I (0) = e^{0} \cos 0 + C = 1 + C = 1 ⟹ C = 0$ Thus, $I (x) = e^{\sin^{2} x} \cos x$ .

06Execute

Evaluate at $x = \frac{π}{3}$ : $I (\frac{π}{3}) = e^{\sin^{2} (π / 3)} \cos (\frac{π}{3}) = e^{(\sqrt{3} / 2)^{2}} \cdot \frac{1}{2} = \frac{1}{2} e^{3 / 4}$

✓Verify

Differentiating $e^{\sin^{2} x} \cos x$ gives $e^{\sin^{2} x} \sin 2 x \cos x - e^{\sin^{2} x} \sin x$ , which matches the integrand identically.

Hints that build this answer step by step

How can the integrand $e^{\sin^{2} x} (\cos x \sin 2 x - \sin x)$ be rewritten to identify an exact derivative?
$\cos x \cdot (2 \sin x \cos x e^{\sin^{2} x}) - \sin x \cdot e^{\sin^{2} x} = \cos x \cdot \frac{d}{d x} (e^{\sin^{2} x}) + \frac{d}{d x} (\cos x) \cdot e^{\sin^{2} x}$
What is the antiderivative $I (x)$ given the initial condition $I (0) = 1$ ?
$I (x) = \cos x \cdot e^{\sin^{2} x}$
What is the value of $I (\frac{π}{3})$ ?
$\frac{1}{2} e^{\frac{3}{4}}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why choose

e^{\sin^{2} x} \sin 2 x

as the second function rather than expanding

\sin 2 x = 2 \sin x \cos x

Because the derivative of $\sin^{2} x$ is $2 \sin x \cos x = \sin 2 x$ , making $\int e^{\sin^{2} x} \sin 2 x d x = e^{\sin^{2} x}$ directly integrable.

Answer

By recognizing the integrand as the derivative of $\cos x \cdot e^{\sin^{2} x}$ , we find $I (x) = \cos x \cdot e^{\sin^{2} x}$ , which gives $I (\frac{π}{3}) = \frac{1}{2} e^{\frac{3}{4}}$ .

Why each option works or fails

A: -1/2e^(3/4) - Carrying an extra negative sign through the product rule differentiation or integration steps leads to $- \frac{1}{2} e^{3 / 4}$ . Check the derivative of $\cos x \cdot e^{\sin^{2} x}$ : $\frac{d}{d x} (\cos x) = - \sin x$ , so the negative sign is already correctly accounted for in the integrand.
B: e^(3/4) - Evaluating $\cos (\frac{π}{3})$ incorrectly as $1$ instead of $\frac{1}{2}$ . Recall standard trigonometric values: $\cos (\frac{π}{3}) = \frac{1}{2}$ and $\sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ .
C · correct: 1/2e^(3/4) - This is the correct value obtained from $I (x) = \cos x \cdot e^{\sin^{2} x}$ evaluated at $x = \frac{π}{3}$ . Correctly integrated and evaluated.
D: -e^(3/4) - Combining both a sign error and an incorrect trigonometric evaluation of $\cos (\frac{π}{3}) = - 1$ . Carefully compute both the antiderivative via product rule inspection and the standard trigonometric values at $\frac{π}{3}$ .

Step-by-step solution

Given: $I (x) = \int e^{\sin^{2} x} (\cos x \sin 2 x - \sin x) d x$ with initial condition $I (0) = 1$ .
Goal: Evaluate $I (\frac{π}{3})$ .
Approach: Split the integrand into $I_{1} = \int e^{\sin^{2} x} \sin 2 x \cos x d x$ and $I_{2} = - \int e^{\sin^{2} x} \sin x d x$ . Apply integration by parts on $I_{1}$ taking $\cos x$ as the first function and $e^{\sin^{2} x} \sin 2 x$ as the second function because $\frac{d}{d x} (e^{\sin^{2} x}) = e^{\sin^{2} x} \sin 2 x$ .
Execute: Integrating $I_{1}$ by parts: $\int \cos x \cdot (e^{\sin^{2} x} \sin 2 x) d x = \cos x \cdot e^{\sin^{2} x} - \int (- \sin x) e^{\sin^{2} x} d x = e^{\sin^{2} x} \cos x + \int e^{\sin^{2} x} \sin x d x$ Combining with $I_{2}$ : $I (x) = e^{\sin^{2} x} \cos x + \int e^{\sin^{2} x} \sin x d x - \int e^{\sin^{2} x} \sin x d x = e^{\sin^{2} x} \cos x + C$
Execute: Using $I (0) = 1$ : $I (0) = e^{0} \cos 0 + C = 1 + C = 1 ⟹ C = 0$ Thus, $I (x) = e^{\sin^{2} x} \cos x$ .
Execute: Evaluate at $x = \frac{π}{3}$ : $I (\frac{π}{3}) = e^{\sin^{2} (π / 3)} \cos (\frac{π}{3}) = e^{(\sqrt{3} / 2)^{2}} \cdot \frac{1}{2} = \frac{1}{2} e^{3 / 4}$
Verify: Differentiating $e^{\sin^{2} x} \cos x$ gives $e^{\sin^{2} x} \sin 2 x \cos x - e^{\sin^{2} x} \sin x$ , which matches the integrand identically.

Indefinite Integration Using Integration by Parts and Chain Rule

What feels right?

By recognizing the integrand as the derivative of $\cos x \cdot e^{\sin^{2} x}$ , we find $I (x) = \cos x \cdot e^{\sin^{2} x}$ , which gives $I (\frac{π}{3}) = \frac{1}{2} e^{\frac{3}{4}}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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