The value of the integral int limits_{-pi / 2 | vedclass

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Question

(D) Let $I = \int \limits_{-\pi / 2}^{\pi / 2} \frac{d x}{\left(1+e^{x}\right)\left(\sin ^{6} x+\cos ^{6} x\right)}$.Using the property $\int_{a}^{b}

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) Let $I = \int \limits_{-\pi / 2}^{\pi / 2} \frac{d x}{\left(1+e^{x}ight)\left(\sin ^{6} x+\cos ^{6} xight)}$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $I = \int \limits_{-\pi / 2}^{\pi / 2} \frac{d x}{\left(1+e^{-x}ight)\left(\sin ^{6} x+\cos ^{6} xight)}$.Adding the two expressions for $I$:$2I = \int \limits_{-\pi / 2}^{\pi / 2} \left( \frac{1}{1+e^x} + \frac{1}{1+e^{-x}} ight) \frac{dx}{\sin^6 x + \cos^6 x}$.Since $\frac{1}{1+e^x} + \frac{1}{1+e^{-x}} = \frac{1}{1+e^x} + \frac{e^x}{e^x+1} = 1$,we get:$2I = \int \limits_{-\pi / 2}^{\pi / 2} \frac{dx}{\sin^6 x + \cos^6 x}$.Since the integrand is an even function,$2I = 2 \int \limits_{0}^{\pi / 2} \frac{dx}{\sin^6 x + \cos^6 x}$,so $I = \int \limits_{0}^{\pi / 2} \frac{dx}{\sin^6 x + \cos^6 x}$.Divide numerator and denominator by $\cos^6 x$:$I = \int \limits_{0}^{\pi / 2} \frac{\sec^6 x dx}{	an^6 x + 1} = \int \limits_{0}^{\pi / 2} \frac{(1+	an^2 x)^2 \sec^2 x dx}{	an^6 x + 1}$.Let $	an x = t$,then $\sec^2 x dx = dt$:$I = \int \limits_{0}^{\infty} \frac{(1+t^2)^2}{t^6+1} dt = \int \limits_{0}^{\infty} \frac{1+2t^2+t^4}{t^6+1} dt$.Using partial fractions or standard integral forms,the value evaluates to $\frac{\pi}{2}$.

The value of the integral $\int \limits_{-\pi / 2}^{\pi / 2} \frac{d x}{\left(1+e^{x}\right)\left(\sin ^{6} x+\cos ^{6} x\right)}$ is equal to

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