MCQMediumJEE 2023Indefinite Integrals

JEE Mathematics 2023 Question with Solution

If I(x)=esin2xcosx(sin2xsinx)dxI(x) = \int e^{\sin^2x}\cos x(\sin 2x - \sin x) \, dx and I(0)=1I(0) = 1, then I(π3)I\left(\frac{\pi}{3}\right) is equal to:

  • A

    e3/4e^{3/4}

  • B

    e3/4-e^{3/4}

  • C

    12e3/4\frac{1}{2}e^{3/4}

  • D

    12e3/4-\frac{1}{2}e^{3/4}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: I(x)=esin2xcosx(sin2xsinx)dxI(x) = \int e^{\sin^2x}\cos x(\sin 2x - \sin x) \, dx and I(0)=1I(0) = 1.

Find: I(π3)I\left(\frac{\pi}{3}\right).

Use sin2x=2sinxcosx\sin 2x = 2\sin x \cos x to rewrite the integrand:

I(x)=esin2xcosx(2sinxcosxsinx)dxI(x)=\int e^{\sin^2x}\cos x\,(2\sin x\cos x-\sin x)\,dx =esin2xsinxcosx(2cosx1)dx=\int e^{\sin^2x}\sin x\cos x\,(2\cos x-1)\,dx

Observe that

ddx(esin2xcosx)=esin2x(2sinxcosx)(cosx)esin2xsinx\frac{d}{dx}\left(e^{\sin^2x}\cos x\right)=e^{\sin^2x}(2\sin x\cos x)(\cos x)-e^{\sin^2x}\sin x =esin2xcosx(2sinxcosxsinx)=e^{\sin^2x}\cos x(2\sin x\cos x-\sin x)

which is exactly the given integrand.

Therefore,

I(x)=esin2xcosx+CI(x)=e^{\sin^2x}\cos x+C

Now apply the initial condition I(0)=1I(0)=1:

1=esin20cos0+C=11+C1=e^{\sin^2 0}\cos 0+C=1\cdot 1+C

So,

C=0C=0

Hence,

I(x)=esin2xcosxI(x)=e^{\sin^2x}\cos x

Evaluate at x=π3x=\frac{\pi}{3}:

I(π3)=esin2(π3)cos(π3)I\left(\frac{\pi}{3}\right)=e^{\sin^2\left(\frac{\pi}{3}\right)}\cos\left(\frac{\pi}{3}\right)

Using sinπ3=32\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} and cosπ3=12\cos \frac{\pi}{3}=\frac{1}{2},

I(π3)=e(32)212=e3412=e342I\left(\frac{\pi}{3}\right)=e^{\left(\frac{\sqrt{3}}{2}\right)^2}\cdot \frac{1}{2}=e^{\frac{3}{4}}\cdot \frac{1}{2}= \frac{e^{\frac{3}{4}}}{2}

Therefore, the correct option is C.

Answer Discrepancy Note

The solution heading states The Correct Option is D, but the worked steps conclude

I(π3)=e342I\left(\frac{\pi}{3}\right)=\frac{e^{\frac{3}{4}}}{2}

which matches option C exactly. The final computed value is therefore taken as authoritative, so the answer is C.

Common mistakes

  • Using the heading Option D without checking the worked solution is incorrect because the actual computation gives e3/42\frac{e^{3/4}}{2}. Always trust the derived result over a mismatched label.

  • Expanding sin2x\sin 2x incorrectly is a common error. The correct identity is sin2x=2sinxcosx\sin 2x = 2\sin x \cos x, and using a wrong identity changes the integrand completely.

  • Forgetting to apply the initial condition I(0)=1I(0)=1 leaves an arbitrary constant CC. After finding the antiderivative, always substitute the given value to determine CC.

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