Find:
\int \frac{\sin 2x}{\left ( \sin ^{2}x+1 \right )\left ( \sin ^{2}x+3 \right )}dx

 

 

 

 
 
 
 
 

Answers (1)

I= \int \frac{\sin 2x}{\left ( \sin ^{2}x +1\right )\left ( \sin ^{2}x+3 \right )}dx
I= \int \frac{2\sin x\cos x}{\left ( \sin ^{2}x+1 \right )\left ( \sin ^{2}x+3 \right )}dx
Let \sin ^{2}x+3= t\: \: \Rightarrow 2\sin x\cos x\, dx= dt
\therefore I= \int \frac{dt}{\left ( t-2 \right )t}
\Rightarrow I= \frac{1}{2}\int \left ( \frac{1}{\left ( t-2 \right )}-\frac{1}{t} \right )dt
I= \frac{1}{2}\left [ \log \left ( t-2 \right ) -\log t\right ]+c
\Rightarrow \frac{1}{2}\left [ \log \left ( \frac{t-2}{t} \right ) \right ]+c
\Rightarrow \log \sqrt{\frac{t-2}{t}}+c                                 \because \frac{1}{n}\log x= \log \left ( x^{\frac{1}{n}} \right )
I\Rightarrow \log \sqrt{\frac{\sin ^{2}x+1}{\sin ^{2}x+3}}+c

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