The value of the definite integral int_{-frac | vedclass

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(C) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)} \cdots(1)$Using the pro

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$\frac{\pi}{2 \sqrt{2}}$

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(C) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}ight)\left(\sin ^{4} x+\cos ^{4} xight)} \cdots(1)$Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,where $a = -\frac{\pi}{4}$ and $b = \frac{\pi}{4}$,we get $a+b = 0$. Thus,$f(x)$ becomes $f(-x)$:$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{-x \cos(-x)}ight)\left(\sin ^{4}(-x)+\cos ^{4}(-x)ight)} = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{-x \cos x}ight)\left(\sin ^{4} x+\cos ^{4} xight)} \cdots(2)$Adding $(1)$ and $(2)$:$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{1}{1+e^{x \cos x}} + \frac{1}{1+e^{-x \cos x}} ight) \frac{d x}{\sin ^{4} x+\cos ^{4} x}$Since $\frac{1}{1+e^{x \cos x}} + \frac{1}{1+e^{-x \cos x}} = \frac{1}{1+e^{x \cos x}} + \frac{e^{x \cos x}}{e^{x \cos x}+1} = 1$,we have:$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x} = 2 \int_{0}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x}$$I = \int_{0}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x} = \int_{0}^{\frac{\pi}{4}} \frac{\sec^4 x}{	an^4 x + 1} dx = \int_{0}^{\frac{\pi}{4}} \frac{(1+	an^2 x) \sec^2 x}{	an^4 x + 1} dx$Let $	an x = u$,then $\sec^2 x dx = du$. As $x 	o 0, u 	o 0$ and $x 	o \frac{\pi}{4}, u 	o 1$:$I = \int_{0}^{1} \frac{1+u^2}{u^4+1} du = \int_{0}^{1} \frac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}} du = \int_{0}^{1} \frac{1+\frac{1}{u^2}}{(u-\frac{1}{u})^2 + 2} du$Let $u-\frac{1}{u} = t$,then $(1+\frac{1}{u^2}) du = dt$. As $u 	o 0, t 	o -\infty$ and $u 	o 1, t 	o 0$:$I = \int_{-\infty}^{0} \frac{dt}{t^2+2} = \left[ \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{t}{\sqrt{2}} ight) ight]_{-\infty}^{0} = \frac{1}{\sqrt{2}} (0 - (-\frac{\pi}{2})) = \frac{\pi}{2\sqrt{2}}$

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

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