Evaluate the definite integral: int_{frac{pi} | vedclass

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(C) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{	an x}}$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,whe

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$\frac{\pi}{12}$

Vedclass · Answer

(C) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{	an x}}$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$,we have $a+b = \frac{\pi}{2}$.$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{	an(\frac{\pi}{2}-x)}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\cot x}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\frac{1}{\sqrt{	an x}}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{	an x}}{\sqrt{	an x}+1} dx$.Adding the two expressions for $I$:$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{1}{1+\sqrt{	an x}} + \frac{\sqrt{	an x}}{1+\sqrt{	an x}} ight) dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$.$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.Therefore,$I = \frac{\pi}{12}$.

Evaluate the definite integral: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}$.

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