Question

evaluate integral

$\frac{1}{1 - sinx} \: dx$
​

Answer 1

Given to find,

$\displaystyle\int\dfrac {1}{1-\sin x}\ dx$

First I multiply both the numerator and the denominator of the fraction by $1+\sin x.$ Then we get,

$\displaystyle\int\dfrac {1}{1-\sin x}\ dx=\int\dfrac {1+\sin x}{(1-\sin x)(1+\sin x)}\ dx\\\\\\\int\dfrac {1}{1-\sin x}\ dx=\int\dfrac {1+\sin x}{1-\sin^2x}\ dx\\\\\\\int\dfrac {1}{1-\sin x}\ dx=\int\dfrac {1+\sin x}{\cos^2x}\ dx\\\\\\\int\dfrac {1}{1-\sin x}\ dx=\int\left (\dfrac {1}{\cos^2x}+\dfrac {\sin x}{\cos^2x}\right)\ dx\\\\\\\int\dfrac {1}{1-\sin x}\ dx=\int\dfrac {1}{\cos^2x}\ dx+\int\dfrac {\tan x}{\cos x}\ dx\\\\\\\int\dfrac {1}{1-\sin x}\ dx=\int\sec^2x\ dx+\int\sec x\tan x\ dx$

But we know that,

$\displaystyle\int\sec^2x\ dx=\tan x+c$

and,

$\displaystyle\int\sec x\tan x\ dx=\sec x+c$

Then, finally,

$\displaystyle\large\boxed {\int\dfrac {1}{1-\sin x}\ dx=\sec x+\tan x+c}$

or if further moved, we get,

$\displaystyle\large\boxed {\int\dfrac {1}{1-\sin x}\ dx=\dfrac {\cos x}{1-\sin x}+c}$

#answerwithquality

#BAL