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Evaluate : $ \int_{\large\frac{\pi}{6}}^{\large\frac{\pi}{3}} \large\frac{dx}{1+\sqrt{tan\:x}} $

1 Answer

  • $\int \limits_a^b f(x)dx=F(b)-F(a)$
  • $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a+b-x) dx$
Step 1:
Given $I=\int \limits _{\pi/6}^{\pi/3} \large\frac{dx}{1+\sqrt {\tan x}}$-----(1)
This can be written as
$I=\int \limits _{\pi/6}^{\pi/3} \large\frac{dx}{1+\sqrt {\sin x}}=\int \limits _{\pi/6}^{\pi/3} \large\frac{\sqrt \cos x}{\sqrt {\cos x}+\sqrt {\sin x}}$$dx$
Applying the property
$ \int \limits_a^b f(x)dx=\int \limits_a^b f(a+b-x) dx$
$I=\int \limits _{\pi/6}^{\pi/3} \large\frac{\sqrt {\cos (\pi/6+\pi/3-x)}}{\sqrt {\cos (\pi/6+\pi/3-x)}+\sin \sqrt {(\pi/6+\pi/3-x)}}$$dx$
$I=\int \limits _{\pi/6}^{\pi/3} \large\frac{\sqrt {\cos (\pi/2-x)}}{\sqrt {\cos (\pi/2-x)}+\sqrt {(\pi/2-x)}}$$dx$
But $\cos (\pi/2-x)=\sin x$ and $ \sin (\pi/2-x)=\cos x$
Therefore $I=\int \limits _{\pi/6}^{\pi/3}\large \frac{\sqrt {\sin x}}{\sqrt {\sin x}+\sqrt {\cos x}}$$dx$-----(2)
Step 2:
Adding equ(1) and equ (2)
$I=\int \limits _{\pi/6}^{\pi/3} \large\frac{\sqrt {\sin x}+\sqrt {\cos x }}{\sqrt {\sin x}+\sqrt {\cos x}}$$dx=\int \limits _{\pi/6}^{\pi/3} dx $
on integrating we get,
$2I=\int \limits _{\pi/6}^{\pi/3} dx$
$\quad=[x]_{\pi/6}^{\pi/3} $
Applying the limits we get
Therefore $I=\large\frac{\pi}{12}$
answered Nov 8, 2013 by sreemathi.v