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

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(B) $I = \int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin x+\cos x} d x \quad \dots(1)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we get:

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$\frac{\pi-1}{4}$

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(B) $I = \int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin x+\cos x} d x \quad \dots(1)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we get:$I = \int_0^{\frac{\pi}{2}} \frac{\cos ^3(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x)+\cos(\frac{\pi}{2}-x)} dx = \int_0^{\frac{\pi}{2}} \frac{\sin ^3 x}{\cos x+\sin x} dx \quad \dots(2)$Adding $(1)$ and $(2)$:$2I = \int_0^{\frac{\pi}{2}} \frac{\sin ^3 x + \cos ^3 x}{\sin x+\cos x} dx$Using the identity $a^3+b^3 = (a+b)(a^2+b^2-ab)$:$2I = \int_0^{\frac{\pi}{2}} \frac{(\sin x+\cos x)(\sin^2 x+\cos^2 x - \sin x \cos x)}{\sin x+\cos x} dx$$2I = \int_0^{\frac{\pi}{2}} (1 - \sin x \cos x) dx$$2I = \int_0^{\frac{\pi}{2}} 1 dx - \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2x) dx$$2I = [x]_0^{\frac{\pi}{2}} - \frac{1}{2} [-\frac{\cos(2x)}{2}]_0^{\frac{\pi}{2}}$$2I = \frac{\pi}{2} + \frac{1}{4} [\cos(\pi) - \cos(0)]$$2I = \frac{\pi}{2} + \frac{1}{4} [-1 - 1] = \frac{\pi}{2} - \frac{2}{4} = \frac{\pi-1}{2}$$I = \frac{\pi-1}{4}$

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

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