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

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

Question

(D) Let $I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{	an x}} dx$ $\qquad ....(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:$I = \int

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) Let $I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{	an x}} dx$ $\qquad ....(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:$I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{	an(\frac{\pi}{2}-x)}} dx$$I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{\cot x}} dx = \int_0^{\frac{\pi}{2}} \frac{\sqrt{	an x}}{1+\sqrt{	an x}} dx$ $\qquad ....(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^{\frac{\pi}{2}} \frac{1+\sqrt{	an x}}{1+\sqrt{	an x}} dx$$2I = \int_0^{\frac{\pi}{2}} 1 dx = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$$I = \frac{\pi}{4}$

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

Similar Questions

$\int_0^{\frac{\pi}{2}} \frac{300 \sin x+100 \cos x}{\sin x+\cos x} \,dx = \ldots$ (in $\pi$)

$\int_{0}^{\pi} \cos^3 x \, dx = $

Suppose $M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} dx$. Then, the value of $(M - N)$ equals

Value of the definite integral $\int_{-1/2}^{1/2} (\sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)) dx$

$\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module