The value of int_{0}^{pi / 2} frac{sin^3 x}{s | vedclass

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(C) Let $I = \int_{0}^{\pi / 2} \frac{\sin^3 x}{\sin x + \cos x} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I =

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

Vedclass · Answer

(C) Let $I = \int_{0}^{\pi / 2} \frac{\sin^3 x}{\sin x + \cos x} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{\pi / 2} \frac{\cos^3 x}{\cos x + \sin x} dx$. Adding the two expressions for $I$: $2I = \int_{0}^{\pi / 2} \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} dx$. Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$: $2I = \int_{0}^{\pi / 2} \frac{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{\sin x + \cos x} dx$. $2I = \int_{0}^{\pi / 2} (1 - \sin x \cos x) dx$. $2I = \int_{0}^{\pi / 2} (1 - \frac{1}{2} \sin 2x) dx$. $2I = [x + \frac{1}{4} \cos 2x]_{0}^{\pi / 2}$. $2I = (\frac{\pi}{2} + \frac{1}{4} \cos \pi) - (0 + \frac{1}{4} \cos 0)$. $2I = (\frac{\pi}{2} - \frac{1}{4}) - (0 + \frac{1}{4}) = \frac{\pi}{2} - \frac{1}{2}$. $I = \frac{\pi - 1}{4}$.

The value of $\int_{0}^{\pi / 2} \frac{\sin^3 x}{\sin x + \cos x} dx$ is

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