# Question bfd5a

Feb 24, 2017

$I = - a r c \tan \left\{\frac{1}{2} \left(\sqrt{\tan} x + \sqrt{\cot} x\right)\right\} + C .$

#### Explanation:

Let, $I = \int \frac{\sqrt{\cot} x - \sqrt{\tan} x}{1 + 3 \sin 2 x} \mathrm{dx} .$

$\therefore I = \int \frac{1 - \tan x}{\sqrt{\tan} x \left(1 + 3 \sin 2 x\right)} \mathrm{dx} .$

Using, $\sin 2 x = \frac{2 \tan x}{1 + {\tan}^{2} x} ,$ we have,

$1 + 3 \sin 2 x = 1 + \frac{6 \tan x}{1 + {\tan}^{2} x} = \frac{{\tan}^{2} x + 6 \tan x + 1}{1 + {\tan}^{2} x} , s o ,$

$I = \int \frac{\left(1 - \tan x\right) \left(1 + {\tan}^{2} x\right)}{\sqrt{\tan} x \left({\tan}^{2} x + 6 \tan x + 1\right)} \mathrm{dx} , \mathmr{and} ,$

$I = \int \frac{\left(1 - \tan x\right) \left({\sec}^{2} x \mathrm{dx}\right)}{\sqrt{\tan} x \left({\tan}^{2} x + 6 \tan x + 1\right)} .$

Now, we use the substn. $\tan x = {t}^{2} \Rightarrow {\sec}^{2} x \mathrm{dx} = 2 t \mathrm{dt} .$

$\therefore I = \int \frac{\left(1 - {t}^{2}\right) \left(2 t \mathrm{dt}\right)}{t \left({t}^{4} + 6 {t}^{2} + 1\right)} = 2 \int \frac{1 - {t}^{2}}{{t}^{4} + 6 {t}^{2} + 1} \mathrm{dt} .$

:. I=2int{t^2(1/t^2-1)}/{t^2(t^2+6+1/t^2)dt#

$= - 2 \int \frac{1 - \frac{1}{t} ^ 2}{{\left(t + \frac{1}{t}\right)}^{2} + 4} \mathrm{dt} .$

Finally, sub.ing, $t + \frac{1}{t} = u \Rightarrow \left(1 - \frac{1}{t} ^ 2\right) \mathrm{dt} = \mathrm{du} .$

$I = - 2 \int \frac{1}{{u}^{2} + 4} \mathrm{du} .$

$= - 2 \left(\frac{1}{2}\right) a r c \tan \left(\frac{u}{2}\right) = - a r c \tan \left(\frac{u}{2}\right) .$$= - a r c \tan \left\{\frac{1}{2} \left(t + \frac{1}{t}\right)\right\}$

$\therefore I = - a r c \tan \left\{\frac{1}{2} \left(\sqrt{\tan} x + \sqrt{\cot} x\right)\right\} + C .$

Enjoy Maths.!