int_0^{pi /2} frac{sqrt{cot x}}{sqrt{cot x} + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{	an x}} \, dx$ ..... $(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$I = \int_0^{\pi /2} \frac{\sqrt{\cot(\pi/2 - x)}}{\sqrt{\cot(\pi/2 - x)} + \sqrt{	an(\pi/2 - x)}} \, dx$Since $\cot(\pi/2 - x) = 	an x$ and $	an(\pi/2 - x) = \cot x$,we have:$I = \int_0^{\pi /2} \frac{\sqrt{	an x}}{\sqrt{	an x} + \sqrt{\cot x}} \, dx$ ..... $(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^{\pi /2} \frac{\sqrt{\cot x} + \sqrt{	an x}}{\sqrt{\cot x} + \sqrt{	an x}} \, dx$$2I = \int_0^{\pi /2} 1 \, dx$$2I = [x]_0^{\pi /2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$I = \frac{\pi}{4}$

$\int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx = $

Similar Questions

$\int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x = \int_0^1 \frac{\log (1+x)}{1+x^2} d x + \int_0^1 f(x) d x$,then $f(x) =$

$\int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$ is equal to

$\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=$

For $n \in N$,let $P_n = \int_1^e (\ln x)^n dx$. Then $(P_{10} - 90P_8)$ is equal to

$\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} dx = . . . . . .$

Vedclass Test Series

Exam Paper Generator

Online Exam Module