int_0^{pi / 2} frac{1}{1+tan ^{2020}(x)} d x= | vedclass

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(C) Let $I = \int_0^{\pi / 2} \frac{1}{1+	an ^{2020} x} d x$. Since $	an x = \frac{\sin x}{\cos x}$,we can write $I = \int_0^{\pi / 2} \frac{\cos ^{

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi / 2} \frac{1}{1+	an ^{2020} x} d x$. Since $	an x = \frac{\sin x}{\cos x}$,we can write $I = \int_0^{\pi / 2} \frac{\cos ^{2020} x}{\cos ^{2020} x+\sin ^{2020} x} d x$ $(i)$. Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get $I = \int_0^{\pi / 2} \frac{\sin ^{2020} x}{\sin ^{2020} x+\cos ^{2020} x} d x$ (ii). Adding equations $(i)$ and (ii),we get $2I = \int_0^{\pi / 2} \frac{\cos ^{2020} x + \sin ^{2020} x}{\cos ^{2020} x + \sin ^{2020} x} d x = \int_0^{\pi / 2} 1 d x$. Thus,$2I = [x]_0^{\pi / 2} = \frac{\pi}{2}$,which implies $I = \frac{\pi}{4}$. Therefore,option $C$ is correct.

$\int_0^{\pi / 2} \frac{1}{1+\tan ^{2020}(x)} d x=$

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