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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Show that $ \int_0^{\large\frac{\pi}{2}} \sqrt{\tan\: x}+\sqrt{\cot x}=\sqrt {2\pi} $

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Toolbox:
  • $\int \limits_a^b f(x)dx=F(b)-f(a)$
  • $\int f(x)dx,if \;f(x)=t\;then\; f'(x)dx=dt$
  • $\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 $x=\large\frac{\pi}{2}$
$\sin\large\frac{\pi}{2}-$$\cos\large\frac{\pi}{2}$$=1$
$\therefore I=\sqrt 2\int_{-1}^1\large\frac{dt}{\sqrt{1-t^2}}$
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 $\sin^{-1}(1)=\large\frac{\pi}{2}$
$\sin^{-1}(-1)=\large\frac{-\pi}{2}$
$\therefore I=\sqrt 2\big[\large\frac{\pi}{2}+\frac{\pi}{2}\big]$
$\qquad=\sqrt 2\pi$
answered Sep 24, 2013 by sreemathi.v
 

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