int_0^{pi /2} {frac{1}{{1 + sqrt {tan x} }}} | vedclass

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

Question

(B) Let $I = \int_0^{\pi /2} \frac{1}{1 + \sqrt{	an x}} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we have: $I = \int_0^{\

Vedclass · Accepted Answer

$\frac{\pi }{4}$

Vedclass · Answer

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

$\int_0^{\pi /2} {\frac{1}{{1 + \sqrt {\tan x} }}} \,dx = $

Similar Questions

$\int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta$ is equal to

The value of the integral $\int_{-1}^{1} \frac{d}{dx} \left( \tan^{-1} \frac{1}{x} \right) dx$ is

$\frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx =$

For $x > 0$,let $f(x) = \int_{1}^{x} \frac{\log t}{1+t} dt$. Then $f(x) + f\left(\frac{1}{x}\right)$ is equal to:

The value of $\int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module