int_0^{pi / 2} frac{d x}{1+tan ^3 x} is equal | vedclass

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(C) Let $I = \int_0^{\pi / 2} \frac{d x}{1+	an ^3 x}$.We know that $	an x = \frac{\sin x}{\cos x}$, so $I = \int_0^{\pi / 2} \frac{\cos ^3 x}{\sin ^

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$\frac{\pi}{4}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi / 2} \frac{d x}{1+	an ^3 x}$.We know that $	an x = \frac{\sin x}{\cos x}$, so $I = \int_0^{\pi / 2} \frac{\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...$(i)$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{\cos ^3(\pi/2 - x)}{\sin ^3(\pi/2 - x) + \cos ^3(\pi/2 - x)} d x$$I = \int_0^{\pi / 2} \frac{\sin ^3 x}{\cos ^3 x + \sin ^3 x} d x$ ...(ii)Adding equations $(i)$ and (ii):$2I = \int_0^{\pi / 2} \frac{\cos ^3 x + \sin ^3 x}{\sin ^3 x + \cos ^3 x} d x$$2I = \int_0^{\pi / 2} 1 d x = [x]_0^{\pi / 2} = \frac{\pi}{2}$Therefore, $I = \frac{\pi}{4}$.

$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is equal to :

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