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Home  >>  CBSE XII  >>  Math  >>  Integrals

$\begin{align*} \int_{\frac{\pi}{6}} ^\frac{\pi}{3} \frac{dx}{1+\sqrt{\tan x}} \end{align*}$


$(A)\; \pi/6$
$(B)\; 2\pi/3$
$(C)\; \pi/12$
$(D)\; \pi/3$

1 Answer

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Let $ I =\begin{align*} \int_{\frac{\pi}{6}} ^\frac{\pi}{3} \frac{dx}{1+\sqrt{\tan x}} = \int_\frac{\pi}{6} ^ \frac{\pi}{3} \frac{\sqrt {\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \end{align*}$
Applying properties :
$ \begin{align*}\int_a^b f(x) dx = \int_a^b f(a+b -x) dx \end{align*}$
$I = \begin{align*} \int_{\pi/6} ^ {\pi/3} \frac{\sqrt{\cos (\pi/3 + \pi/6 - x)}} { \sqrt{\cos (\pi/3 + \pi/6 - x )} + \sqrt{\sin(\pi/3 + \pi/6 - x)}} dx \end{align*}$
$I = \begin{align*} \int_{\pi/6} ^ {\pi/3} \frac{\sqrt{\cos (\pi/2- x)}} { \sqrt{\cos (\pi/2 - x )} + \sqrt{\sin(\pi/2- x)}} dx \end{align*}$
$\begin{align*}I =\int_{\pi/6} ^{\pi/3} \frac{\sqrt{\sin x} }{ \sqrt{\sin x} + \sqrt{\cos x}} dx \end{align*} - - - -(2)$
Adding eq (1) and (2)
$\begin{align*} 2I = \int _{\pi/6}{\pi/3} \frac{\sqrt{\cos x + \sin x}}{\sqrt {\sin x + \sin x}} dx \end{align*}$
$\begin{align*}2I &= \int_{\pi/3} ^ {\pi/6} dx \\ \implies 2I & = [x ]_{\pi/6} ^ {\pi/3} \end{align*}$
$2I = \pi/3 - \pi/6 =\pi/6$
$\implies I = \pi/ 12$
answered Dec 12, 2016 by priyanka.c
 
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