The value of the integral is: [2026]

Question

The value of the integral $\int_{- \frac{π}{4}}^{\frac{π}{4}} (\frac{32 \cos^{4} x}{1 + e^{\sin x}}) d x$ is: [2026]

Smart Achievers · Accepted Answer

(2)

Let $I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{32 \cos^{4} x}{1 + e^{\sin x}} d x$ ... (i)

$\Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{(32 \cos^{4} x)}{e^{\sin x} + 1} e^{\sin x} d x$ ... (ii)

Adding equations (i) and (ii), we get

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

$\Rightarrow I = 16 \times 2 \int_{0}^{\frac{π}{4}} {(\cos^{2} x)}^{2} d x$

$= 32 \int_{0}^{\frac{π}{4}} {(\frac{1 + \cos 2 x}{2})}^{2} d x$

$= 8 \int_{0}^{\frac{π}{4}} (1 + \cos^{2} 2 x + 2 \cos 2 x) d x$

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

$= 8 \times \frac{π}{4} + 8 ((\frac{1}{2} \times \frac{π}{4}) + \frac{1}{2} {(\frac{\sin 4 x}{4})}_{0}^{π / 4}) + 8$

$= 2 π + π + 0 + 8 = 3 π + 8$

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