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

 

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