The value of the integral int_{-pi/4}^{pi/4} | vedclass

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Question

(B) Let $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x}{1 + e^{\sin x}} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a+b=0$,

Vedclass · Accepted Answer

$3\pi + 8$

Vedclass · Answer

(B) Let $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x}{1 + e^{\sin x}} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a+b=0$,we have $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 (-x)}{1 + e^{\sin (-x)}} dx = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x}{1 + e^{-\sin x}} dx$. $I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x e^{\sin x}}{e^{\sin x} + 1} dx$. Adding the two expressions for $I$: $2I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 x (1 + e^{\sin x})}{1 + e^{\sin x}} dx = \int_{-\pi/4}^{\pi/4} 32 \cos^4 x dx$. Since $\cos^4 x$ is an even function,$2I = 64 \int_0^{\pi/4} \cos^4 x dx$. Using the identity $\cos^4 x = \frac{1}{8}(3 + 4\cos 2x + \cos 4x)$: $I = 32 \int_0^{\pi/4} \frac{3 + 4\cos 2x + \cos 4x}{8} dx = 4 \int_0^{\pi/4} (3 + 4\cos 2x + \cos 4x) dx$. $I = 4 [3x + 2\sin 2x + \frac{1}{4}\sin 4x]_0^{\pi/4} = 4(3(\pi/4) + 2\sin(\pi/2) + \frac{1}{4}\sin(\pi)) - 4(0)$. $I = 4(3\pi/4 + 2(1) + 0) = 3\pi + 8$.

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

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