The value of the integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{32 \cos ^{4} x}{1+e^{\sin x}}\right) d x$ is :

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Definite Integrals

The value of the integral

\int_{- \frac{π}{4}}^{\frac{π}{4}} (\frac{32 c o s ^{4} x}{1 + e ^{s i n x}}) d x

is :

Answer: (B)

Let

I = \int_{- π /4}^{π /4} \frac{32 c o s ^{4} x}{1 + e ^{s i n x}} d x

. Replacing

x

- x

gives

I = \int_{- π /4}^{π /4} \frac{32 c o s ^{4} x}{1 + e ^{- s i n x}} d x

. Adding the two forms,

2 I = \int_{- π /4}^{π /4} 32 cos^{4} x d x

, so

I = \int_{0}^{π /4} 32 cos^{4} x d x

. Using

cos^{4} x = \frac{3 + 4 c o s 2 x + c o s 4 x}{8}

I = 4 \int_{0}^{π /4} (3 + 4 cos 2 x + cos 4 x) d x = 4 [3 x + 2 sin 2 x + \frac{1}{4} sin 4 x]_{0}^{π /4} = 4 (\frac{3 π}{4} + 2) = 3 π + 8

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