Definite IntegrationmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

King-Property Integral = 3π + 8 | JEE Main 2026

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

Question

The value of the integral

\displaystyle\int_{-\pi/4}^{\pi/4}\left(\dfrac{32\cos^4 x}{1+e^{\sin x}}\right)dx

4\pi+2

3\pi+8

correct

3\pi+4

4\pi+3

Solution

Step 1: Let

I=\displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{32\cos^4 x}{1+e^{\sin x}}dx

. Replace

x\to-x

(

\cos^4

even,

\sin(-x)=-\sin x

I=\int_{-\pi/4}^{\pi/4}\dfrac{32\cos^4 x}{1+e^{-\sin x}}dx.

Add:

2I=\int_{-\pi/4}^{\pi/4}32\cos^4 x\left(\dfrac{1}{1+e^{\sin x}}+\dfrac{1}{1+e^{-\sin x}}\right)dx.

\dfrac{1}{1+e^{u}}+\dfrac{1}{1+e^{-u}}=\dfrac{1}{1+e^{u}}+\dfrac{e^{u}}{e^{u}+1}=1.

\Rightarrow2I=\int_{-\pi/4}^{\pi/4}32\cos^4 x\,dx\Rightarrow I=\int_{0}^{\pi/4}32\cos^4 x\,dx.

Step 2:

\cos^2\theta=\dfrac{1+\cos2\theta}{2}

\cos^4 x=\left(\dfrac{1+\cos2x}{2}\right)^2=\dfrac{1+2\cos2x+\cos^2 2x}{4}=\dfrac{1}{4}\left(1+2\cos2x+\dfrac{1+\cos4x}{2}\right).

\Rightarrow32\cos^4 x=8\left(1+2\cos2x+\dfrac{1+\cos4x}{2}\right).

I=8\int_0^{\pi/4}\left(\dfrac{3}{2}+2\cos2x+\dfrac{\cos4x}{2}\right)dx.

Step 3:

I=8\left[\dfrac{3x}{2}+\sin2x+\dfrac{\sin4x}{8}\right]_0^{\pi/4}.

x=\dfrac{\pi}{4}

\dfrac{3}{2}\cdot\dfrac{\pi}{4}=\dfrac{3\pi}{8}

\sin\dfrac{\pi}{2}=1

\dfrac{\sin\pi}{8}=0

; at

x=0

0

\therefore I=8\left(\dfrac{3\pi}{8}+1\right)=3\pi+8.

Correct answer: (2)

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