int_0^{3 pi / 2} frac{cos ^3 x}{cos ^3 x+sin | vedclass

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Question

(C) Let $I = \int_0^{3 \pi / 2} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \i

Vedclass · Accepted Answer

$\frac{3 \pi}{4}$

Vedclass · Answer

(C) Let $I = \int_0^{3 \pi / 2} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{3 \pi / 2} \frac{\cos ^3 (3 \pi / 2 - x)}{\cos ^3 (3 \pi / 2 - x) + \sin ^3 (3 \pi / 2 - x)} d x$. Since $\cos(3 \pi / 2 - x) = -\sin x$ and $\sin(3 \pi / 2 - x) = -\cos x$,we have: $I = \int_0^{3 \pi / 2} \frac{(-\sin x)^3}{(-\sin x)^3 + (-\cos x)^3} d x = \int_0^{3 \pi / 2} \frac{-\sin ^3 x}{-\sin ^3 x - \cos ^3 x} d x = \int_0^{3 \pi / 2} \frac{\sin ^3 x}{\sin ^3 x + \cos ^3 x} d x$. Adding the two expressions for $I$: $2I = \int_0^{3 \pi / 2} \frac{\cos ^3 x + \sin ^3 x}{\cos ^3 x + \sin ^3 x} d x = \int_0^{3 \pi / 2} 1 d x$. $2I = [x]_0^{3 \pi / 2} = \frac{3 \pi}{2}$. Therefore,$I = \frac{3 \pi}{4}$.

$\int_0^{3 \pi / 2} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x=$

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