Browse Questions

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

Can you answer this question?

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$
$\therefore I=\sqrt 2\int_{-1}^1\large\frac{dt}{\sqrt{1-t^2}}$