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Integral of e^(sin²x)(cosx+cos³x)sinx | JEE

JEE Maths question with a full step-by-step solution.

Question
∫esin⁡2x (cos⁡x+cos⁡3x)sin⁡x dx\displaystyle\int e^{\sin^{2}x}\,(\cos x+\cos^{3}x)\sin x\,dx equals
A12esin⁡2x(3−sin⁡2x)+C\dfrac{1}{2}e^{\sin^{2}x}\big(3-\sin^{2}x\big)+Ccorrect
B12esin⁡2x(1−12cos⁡2x)+C\dfrac{1}{2}e^{\sin^{2}x}\left(1-\dfrac{1}{2}\cos^{2}x\right)+C
Cesin⁡2x(3cos⁡2x+2sin⁡2x)+Ce^{\sin^{2}x}\big(3\cos^{2}x+2\sin^{2}x\big)+C
Desin⁡2x(2cos⁡2x+3sin⁡2x)+Ce^{\sin^{2}x}\big(2\cos^{2}x+3\sin^{2}x\big)+C
Solution
Step 1: Simplify the trigonometric factor:
(cos⁡x+cos⁡3x)sin⁡x=cos⁡x(1+cos⁡2x)sin⁡x=(1+cos⁡2x)sin⁡xcos⁡x(\cos x+\cos^{3}x)\sin x=\cos x(1+\cos^{2}x)\sin x=(1+\cos^{2}x)\sin x\cos x
 Step 2: Substitute t=sin⁡2xt=\sin^{2}x. Then cos⁡2x=1−t\cos^{2}x=1-t and dt=2sin⁡xcos⁡x dxdt=2\sin x\cos x\,dx, i.e. sin⁡xcos⁡x dx=dt2\sin x\cos x\,dx=\dfrac{dt}{2}. Step 3: Rewrite the integral:
I=∫esin⁡2x(1+cos⁡2x)sin⁡xcos⁡x dx=∫et(1+(1−t))dt2=12∫et(2−t) dtI=\int e^{\sin^{2}x}(1+\cos^{2}x)\sin x\cos x\,dx=\int e^{t}\big(1+(1-t)\big)\dfrac{dt}{2}=\dfrac{1}{2}\int e^{t}(2-t)\,dt
 Step 4: Integrate by parts, ∫(2−t)et dt=(2−t)et−∫(−1)et dt=(2−t)et+et=(3−t)et\displaystyle\int (2-t)e^{t}\,dt=(2-t)e^{t}-\int(-1)e^{t}\,dt=(2-t)e^{t}+e^{t}=(3-t)e^{t}:
I=12(3−t)et+CI=\dfrac{1}{2}(3-t)e^{t}+C
 Step 5: Back-substitute t=sin⁡2xt=\sin^{2}x:
I=12esin⁡2x(3−sin⁡2x)+CI=\dfrac{1}{2}e^{\sin^{2}x}\big(3-\sin^{2}x\big)+C
 Correct answer: (1)
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