Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Evaluate:\[\int\limits_0^{\pi/2} \frac {\sqrt {\tan x}}{1+\sqrt {\tan x}}dx\]

Can you answer this question?

1 Answer

0 votes
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii) $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a-x) dx$
Given $\int\limits_0^{\pi/2} \frac {\sqrt {\tan x}}{1+\sqrt {\tan x}}dx \qquad But\; \tan x =\frac{\sin x}{\cos x}$
This can be written as
$I=\int \limits_0^{\pi/2} \Large\frac{\frac{\sqrt {\sin x}}{\sqrt {\cos x}}}{1+\sqrt {\frac{\sin x}{\cos x}}}$
$=\int \limits_0 ^ {\pi/2} \large\frac{\sqrt {\sin x}}{\sqrt {\cos x}+\sqrt {\sin x}}$-----(1)
By applying the property $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a-x) dx$
$I=\int \limits_0^{\pi/2} \large\frac{\sqrt {\sin (\pi/2-x)}}{\sqrt {\cos (\pi/2-x)}+\sqrt {\sin (\pi/2-x)}} dx$
But $ \sin (\pi/2-x)=\cos x \;and\; \cos (\pi/2-x)=\sin x$
Therefore $I=\int \limits_0^{\pi/2} \large\frac{\sqrt {\cos x}}{\sqrt {\sin x}+\sqrt {\cos x}}dx$ -----(2)
Adding equ(1) and equ (2)
$2I=\int \limits_0 ^ {\pi/2} \large\frac{\sqrt {\sin x}+\sqrt {\cos x}}{\sqrt {\sin x}+\sqrt {\cos x}}dx=\int \limits_0^{\pi/2} dx $
On integrating we get $2I=\bigg[x \bigg]_0^{\pi/2}$
On applying limits$2I=\pi/2 -0 $
Therefore $2I=\frac{\pi}{2}$



answered Mar 15, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App