Question

integration of (sinx+tanx)/(sinx-tanx)​

Answer 1

Step-by-step explanation:

We have,

$\int \frac{ \sin(x) + \tan(x) }{ \sin(x) - \tan(x) } dx$

$= \int \frac{ \frac{2 \tan( \frac{x}{2} ) }{1 + \tan^{2} ( \frac{x}{2} ) } + \frac{2 \tan( \frac{x}{2} ) }{1 - \tan^{2} ( \frac{x}{2} ) } }{ \frac{2 \tan( \frac{x}{2} ) }{1 + \tan^{2} ( \frac{x}{2} ) } - \frac{2 \tan( \frac{x}{2} ) }{1 - \tan^{2} ( \frac{x}{2} ) } } dx$

$= \int \frac{ \frac{1}{1 + \tan^{2} ( \frac{x}{2} ) } + \frac{1 }{1 - \tan^{2} ( \frac{x}{2} ) } }{ \frac{1}{1 + \tan^{2} ( \frac{x}{2} ) } - \frac{1}{1 - \tan^{2} ( \frac{x}{2} ) } } dx$

$= \int \frac{2}{ - 2 \tan^{2} ( \frac{x}{2} ) } dx$

$= - \int \cot^{2} ( \frac{x}{2} ) dx$

$= - 2\int \frac{1}{2} \cot^{2} ( \frac{x}{2} ) dx$

$= - 2\int \frac{1}{2} ( \cosec^{2} ( \frac{x}{2} ) - 1) dx$

$= - 2\int \frac{1}{2} (\cosec^{2} ( \frac{x}{2} ))dx + 1 \int \: dx$

$= - 2 \cot( \frac{x}{2} ) + x + c$