int_{-pi / 8}^{pi / 8} frac{sin ^4(4 x)}{1+e^ | vedclass

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(C) Let $I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(4 x)}{1+e^{4 x}} d x$ ....$(i)$Using the property $\int_a^b f(x) d x = \int_a^b f(a+b

Vedclass · Accepted Answer

$\frac{3 \pi}{64}$

Vedclass · Answer

(C) Let $I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(4 x)}{1+e^{4 x}} d x$ ....$(i)$Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$, where $a = -\frac{\pi}{8}$ and $b = \frac{\pi}{8}$, we have $a+b = 0$.$I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(-4 x)}{1+e^{-4 x}} d x = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(4 x)}{1+\frac{1}{e^{4 x}}} d x = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{e^{4 x} \sin ^4(4 x)}{e^{4 x}+1} d x$ ....(ii)Adding $(i)$ and (ii):$2I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{(1+e^{4 x}) \sin ^4(4 x)}{1+e^{4 x}} d x = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \sin ^4(4 x) d x$Since $\sin^4(4x)$ is an even function, $2I = 2 \int_0^{\frac{\pi}{8}} \sin ^4(4 x) d x$, so $I = \int_0^{\frac{\pi}{8}} \sin ^4(4 x) d x$.Let $4x = t$, then $4 dx = dt$, so $dx = \frac{1}{4} dt$. When $x=0, t=0$; when $x=\frac{\pi}{8}, t=\frac{\pi}{2}$.$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \sin ^4 t d t$.Using Wallis' formula $\int_0^{\frac{\pi}{2}} \sin^n t dt = \frac{(n-1)!!}{n!!} 	imes \frac{\pi}{2}$ for even $n$:$I = \frac{1}{4} 	imes \left( \frac{3 	imes 1}{4 	imes 2} 	imes \frac{\pi}{2} ight) = \frac{1}{4} 	imes \frac{3 \pi}{16} = \frac{3 \pi}{64}$.

$\int_{-\pi / 8}^{\pi / 8} \frac{\sin ^4(4 x)}{1+e^{4 x}} d x=$

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