Find: <img alt="\int \frac{2\cos \, x}{\left ( 1-\sin \, x \right )\left ( 2-\cos ^{2} x\right )}dx." src="https://entrancecorner.codecogs.com/gif.latex?%5Cint%20%5Cfrac%7B2%5Ccos%20%5C%2C%20x%7D%7B%5Cleft%20%28%201-%5Csin%20%5C%2C%20x%20%5Crigh

Find:
$\int \frac{2\cos \, x}{\left ( 1-\sin \, x \right )\left ( 2-\cos ^{2} x\right )}dx.$

Answers (1)

$I= \int \frac{2\cos x}{\left ( 1-\sin x \right )\left ( 2-\cos ^{2}x \right )}dx$
$\Rightarrow I= \int \frac{2\cos x }{\left ( 1-\sin x \right )\left ( 2-1+\sin ^{2}x \right )}dx$
$\Rightarrow \int \frac{2\cos x\, dx}{\left ( 1-\sin x \right )\left ( 1-\sin ^{2}x \right )}$
Now, Let   $\sin x= t$
$\cos x\, dx= dt$
Then   $I= \int \frac{2dt}{\left ( 1-t \right )\left ( 1+t^{2} \right )}$
Then, solving it by partial fraction
$\Rightarrow \frac{2}{\left ( 1-t \right )\left ( 1+t^{2} \right )}= \frac{A}{\left ( 1-t \right )}+\frac{Bt+C}{\left ( 1+t^{2} \right )}$
$\Rightarrow 2= A\left ( 1+t^{2} \right )+\left ( Bt+C \right )\left ( 1-t \right )$
$\Rightarrow 2= A+At^{2}+Bt-Bt^{2}+C-Ct$
$\Rightarrow 2= t^{2} \times \left ( A-B \right )+t\left ( B-C \right )+A+C$
Equaling the coefficient of $t^{2},t$ and of constant lines of both sides,
$A-B= 0$
$B-C= 0$
   $A+C= 2$
On solving $A= C= 1\: and \: B= 1$
$\therefore \int \frac{2dt}{\left ( 1-t \right )\left ( 1+t^{2} \right )}= \int \left [ \frac{1}{\left ( 1-t \right )} +\left ( \frac{t+1}{1+t^{2}} \right )\right ]$
$I= \int \frac{1}{1-t}dt+\int \frac{tdt}{\left ( 1+t^{2} \right )}+ \int \frac{1}{\left ( 1+t^{2} \right )}dt$
$= \log \left | 1-t \right |+\frac{1}{2}\: \log \left | 1+t^{2} \right |+\tan^{-1}t$
$= \log \left | 1-\sin x \right |+\frac{1}{2}\: \log \left | 1+\sin ^{2}x \right |+\tan^{-1}\left ( \sin x \right )+C$