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

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(C) ધારો કે $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$  $(i)$ ગુણધર્મ $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(C) ધારો કે $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$  $(i)$ ગુણધર્મ $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$ નો ઉપયોગ કરતા: $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)}{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x) + \cos ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)} d x$ $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{2} - x)}{\sin ^{3}(\frac{\pi}{2} - x) + \cos ^{3}(\frac{\pi}{2} - x)} d x$ કારણ કે $\sin(\frac{\pi}{2} - x) = \cos x$ અને $\cos(\frac{\pi}{2} - x) = \sin x$,તેથી: $I = \int_{\pi / 6}^{\pi / 3} \frac{\cos ^{3} x}{\cos ^{3} x + \sin ^{3} x} d x$  (ii) $(i)$ અને (ii) નો સરવાળો કરતા: $2I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x + \cos ^{3} x}{\sin ^{3} x + \cos ^{3} x} d x$ $2I = \int_{\pi / 6}^{\pi / 3} 1 d x$ $2I = [x]_{\pi / 6}^{\pi / 3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$ $I = \frac{\pi}{12}$

$\int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$ ની કિંમત શોધો.

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