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Q31  Integrate the functions \frac{\sin x }{( 1+ \cos x )^2}

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Given function  \frac{\sin x }{( 1+ \cos x )^2},

Assume the 1+\cos x =t

 \therefore -\sin x dx =dt

\implies \int \frac{\sin x}{(1+\cos x)^2}dx = \int -\frac{dt}{t^2}

= -\int t^{-2}dt

= \frac{1}{t}+C

Now, back substituted the value of t.

= \frac{1}{1+\cos x} +C, where C is any constant value.

Posted by

Divya Prakash Singh

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