int_{frac{pi}{6}}^{frac{pi}{3}} frac{sqrt{cos | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(D) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$. Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we have $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$. So,$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+\sqrt{\cos(\frac{\pi}{2}-x)}} d x$. $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$. Adding the two expressions for $I$: $2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x$. $2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$. Therefore,$I = \frac{\pi}{12}$.

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

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