Why choose as the second function rather than expanding ?

Because the derivative of is , making directly integrable.

JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

I(x)=\int e^{\sin^2 x}(\cos x\sin 2x-\sin x)dx

and

I(0)=1

, then

I\left(\frac{\pi}{3}\right)

is equal to

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-\frac{1}{2}e^{\frac{3}{4}}

e^{\frac{3}{4}}

\frac{1}{2}e^{\frac{3}{4}}

-e^{\frac{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 $I\left(\frac{\pi}{3}\right) = \frac{1}{2}e^{\frac{3}{4}}$ .

Option analysis

Why each option works or fails

A ·

-\frac{1}{2}e^{\frac{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 $\cos x \cdot e^{\sin^2 x}$ : $\frac{d}{dx}(\cos x) = -\sin x$ , so the negative sign is already correctly accounted for in the integrand.

B ·

e^{\frac{3}{4}}

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

C ·

\frac{1}{2}e^{\frac{3}{4}}

None. This is the correct value obtained from $I(x) = \cos x \cdot e^{\sin^2 x}$ evaluated at $x = \frac{\pi}{3}$ . Correctly integrated and evaluated.

D ·

-e^{\frac{3}{4}}

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

Reviewed route

Solution

StepWorking

01given

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

02goal

Evaluate $I\left(\frac{\pi}{3}\right)$ .

03approach

Split the integrand into $I_1 = \int e^{\sin^2 x}\sin 2x \cos x\,dx$ and $I_2 = -\int e^{\sin^2 x}\sin x\,dx$ . Apply integration by parts on $I_1$ taking $\cos x$ as the first function and $e^{\sin^2 x}\sin 2x$ as the second function because $\frac{d}{dx}(e^{\sin^2 x}) = e^{\sin^2 x}\sin 2x$ .

04execute

Integrating $I_1$ by parts: $\int \cos x \cdot \left(e^{\sin^2 x}\sin 2x\right)dx = \cos x \cdot e^{\sin^2 x} - \int (-\sin x)e^{\sin^2 x}\,dx = e^{\sin^2 x}\cos x + \int e^{\sin^2 x}\sin x\,dx$ Combining with $I_2$ : $I(x) = e^{\sin^2 x}\cos x + \int e^{\sin^2 x}\sin x\,dx - \int e^{\sin^2 x}\sin x\,dx = e^{\sin^2 x}\cos x + C$

05execute

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

06execute

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

✓verify

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why choose

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

as the second function rather than expanding

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

Because the derivative of $\sin^2 x$ is $2\sin x \cos x = \sin 2x$ , making $\int e^{\sin^2 x}\sin 2x\,dx = e^{\sin^2 x}$ 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 $I\left(\frac{\pi}{3}\right) = \frac{1}{2}e^{\frac{3}{4}}$ .

Why each option works or fails

A: $-\frac{1}{2}e^{\frac{3}{4}}$ - 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 $\cos x \cdot e^{\sin^2 x}$ : $\frac{d}{dx}(\cos x) = -\sin x$ , so the negative sign is already correctly accounted for in the integrand.
B: $e^{\frac{3}{4}}$ - Evaluating $\cos\left(\frac{\pi}{3}\right)$ incorrectly as $1$ instead of $\frac{1}{2}$ . Recall standard trigonometric values: $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ and $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ .
C · correct: $\frac{1}{2}e^{\frac{3}{4}}$ - None. This is the correct value obtained from $I(x) = \cos x \cdot e^{\sin^2 x}$ evaluated at $x = \frac{\pi}{3}$ . Correctly integrated and evaluated.
D: $-e^{\frac{3}{4}}$ - Combining both a sign error and an incorrect trigonometric evaluation of $\cos\left(\frac{\pi}{3}\right) = -1$ . Carefully compute both the antiderivative via product rule inspection and the standard trigonometric values at $\frac{\pi}{3}$ .

Step-by-step solution

given: $I(x) = \int e^{\sin^2 x}(\cos x \sin 2x - \sin x)\,dx$ with initial condition $I(0) = 1$ .
goal: Evaluate $I\left(\frac{\pi}{3}\right)$ .
approach: Split the integrand into $I_1 = \int e^{\sin^2 x}\sin 2x \cos x\,dx$ and $I_2 = -\int e^{\sin^2 x}\sin x\,dx$ . Apply integration by parts on $I_1$ taking $\cos x$ as the first function and $e^{\sin^2 x}\sin 2x$ as the second function because $\frac{d}{dx}(e^{\sin^2 x}) = e^{\sin^2 x}\sin 2x$ .
execute: Integrating $I_1$ by parts: $\int \cos x \cdot \left(e^{\sin^2 x}\sin 2x\right)dx = \cos x \cdot e^{\sin^2 x} - \int (-\sin x)e^{\sin^2 x}\,dx = e^{\sin^2 x}\cos x + \int e^{\sin^2 x}\sin x\,dx$ Combining with $I_2$ : $I(x) = e^{\sin^2 x}\cos x + \int e^{\sin^2 x}\sin x\,dx - \int e^{\sin^2 x}\sin x\,dx = e^{\sin^2 x}\cos x + C$
execute: Using $I(0) = 1$ : $I(0) = e^0 \cos 0 + C = 1 + C = 1 \implies C = 0$ Thus, $I(x) = e^{\sin^2 x}\cos x$ .
execute: Evaluate at $x = \frac{\pi}{3}$ : $I\left(\frac{\pi}{3}\right) = e^{\sin^2(\pi/3)}\cos\left(\frac{\pi}{3}\right) = e^{(\sqrt{3}/2)^2} \cdot \frac{1}{2} = \frac{1}{2}e^{3/4}$
verify: Differentiating $e^{\sin^2 x}\cos x$ gives $e^{\sin^2 x}\sin 2x \cos x - e^{\sin^2 x}\sin x$ , which matches the integrand identically.

Integral Calculus: JEE Main Mathematics Question with Solution

By recognizing the integrand as the derivative of cos⁡x⋅esin⁡2x\cos x \cdot e^{\sin^2 x}cosx⋅esin2x, we find I(x)=cos⁡x⋅esin⁡2xI(x) = \cos x \cdot e^{\sin^2 x}I(x)=cosx⋅esin2x, which gives I(π3)=12e34I\left(\frac{\pi}{3}\right) = \frac{1}{2}e^{\frac{3}{4}}I(3π​)=21​e43​.

01given

02goal

03approach

04execute

05execute

06execute

✓verify

Students also ask

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 $I\left(\frac{\pi}{3}\right) = \frac{1}{2}e^{\frac{3}{4}}$ .