int_{pi / 6}^{pi / 3} frac{sin ^{3} x}{sin ^{ | vedclass

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(C) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$  $(i)$ Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b}

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$\frac{\pi}{12}$

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(C) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$  $(i)$ Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we have: $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)}{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x) + \cos ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)} d x$ $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{2} - x)}{\sin ^{3}(\frac{\pi}{2} - x) + \cos ^{3}(\frac{\pi}{2} - x)} d x$ Since $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$,we get: $I = \int_{\pi / 6}^{\pi / 3} \frac{\cos ^{3} x}{\cos ^{3} x + \sin ^{3} x} d x$  (ii) Adding $(i)$ and (ii): $2I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x + \cos ^{3} x}{\sin ^{3} x + \cos ^{3} x} d x$ $2I = \int_{\pi / 6}^{\pi / 3} 1 d x$ $2I = [x]_{\pi / 6}^{\pi / 3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$ $I = \frac{\pi}{12}$

$\int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$ is equal to

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