Ask Questions, Get Answers

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

Show that $ \int_0^{\large\frac{\pi}{2}} \sqrt{\tan\: x}+\sqrt{\cot x}=\sqrt {2\pi} $

Can you answer this question?

1 Answer

0 votes
  • $\int \limits_a^b f(x)dx=F(b)-f(a)$
  • $\int f(x)dx,if \;f(x)=t\;then\; f'(x)dx=dt$
  • $\int\large\frac{dx}{\sqrt{1-x^2}}$$=\sin^{-1}x+c$
Step 1:
Let $I=\int_0^{\Large\frac{\pi}{2}}\sqrt{ \tan x}+\sqrt{\cot x}dx$
This can be written as
$I=\int_0^{\Large\frac{\pi}{2}}\bigg(\sqrt{\large\frac{\sin x}{\cos x}}+\sqrt{\large\frac{\cos x}{\sin x}}\bigg)$
$\quad=\int_0^{\Large\frac{\pi}{2}}\bigg[\large\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\bigg]$$dx$
Multiply and divide by $\sqrt 2$
$\quad=\sqrt 2\int_0^{\Large\frac{\pi}{2}}\large\frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}$
$\quad=\sqrt 2\int_0^{\Large\frac{\pi}{2}}\large\frac{\sin x+\cos x}{\sqrt{1-(1-2\sin x\cos x)}}$
But $1-2\sin x\cos x=(\sin x-\cos x)^2$
$sin^2 x+\cos^2x=1$
$\therefore I=\sqrt 2\int_0^{\Large\frac{\pi}{2}}\large\frac{(\sin x+\cos x)dx}{\sqrt{1-(\sin x-\cos x)^2}}$
Step 2:
Put $\sin x-\cos x=t$
Differentiating with respect to $ x$
$(\cos x+\sin x)dx=dt$
When $x=0$
$t=\sin 0-\cos 0=-1$
When $x=\large\frac{\pi}{2}$
$\therefore I=\sqrt 2\int_{-1}^1\large\frac{dt}{\sqrt{1-t^2}}$
But $\int\large\frac{dx}{\sqrt{1-x^2}}$$=\sin^{-1}x+c$
$I=\sqrt 2\big[\sin^{-1}t\big]_{-1}^1$
Step 3:
Now applying the limits we get,
$I=\sqrt 2\big[\sin^{-1}(1)-\sin^{-1}(-1)\big]$
But $\sin^{-1}(1)=\large\frac{\pi}{2}$
$\therefore I=\sqrt 2\big[\large\frac{\pi}{2}+\frac{\pi}{2}\big]$
$\qquad=\sqrt 2\pi$
answered Sep 24, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App