The value of the integral int_0^{frac{pi}{2}} | vedclass

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Question

(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{	an x}} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x)

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{	an x}} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$, we get: $I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot(\frac{\pi}{2}-x)}}{\sqrt{\cot(\frac{\pi}{2}-x)}+\sqrt{	an(\frac{\pi}{2}-x)}} \,dx$. Since $\cot(\frac{\pi}{2}-x) = 	an x$ and $	an(\frac{\pi}{2}-x) = \cot x$, we have: $I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{	an x}}{\sqrt{	an x}+\sqrt{\cot x}} \,dx$. Adding the two expressions for $I$: $2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}+\sqrt{	an x}}{\sqrt{\cot x}+\sqrt{	an x}} \,dx = \int_0^{\frac{\pi}{2}} 1 \,dx$. $2I = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$. Therefore, $I = \frac{\pi}{4}$.

The value of the integral $\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} \,dx$ is

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