int_{frac{pi}{18}}^{frac{4pi}{9}} frac{2sqrt{ | vedclass

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Question

(C) Let $I = \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ . . . $(i)$ Using the property $\int_{a}^

Vedclass · Accepted Answer

$\frac{7\pi}{18}$

Vedclass · Answer

(C) Let $I = \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ . . . $(i)$ Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = \frac{\pi}{18}$ and $b = \frac{4\pi}{9}$,we have $a+b = \frac{\pi}{18} + \frac{8\pi}{18} = \frac{9\pi}{18} = \frac{\pi}{2}$. So,$I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx$ $I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ . . . $(ii)$ Adding $(i)$ and $(ii)$: $2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) dx$ $2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} 1 dx = 2 [x]_{\frac{\pi}{18}}^{\frac{4\pi}{9}}$ $2I = 2 \left( \frac{4\pi}{9} - \frac{\pi}{18} \right) = 2 \left( \frac{8\pi - \pi}{18} \right) = 2 \left( \frac{7\pi}{18} \right) = \frac{7\pi}{9}$ $I = \frac{7\pi}{18}$

$\int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = . . . . . .$

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