Find: 2 (1- )(1+ ^2x) dx - Vidyadip

Question

Find: $\int\frac{2\cos\text{x}}{(1-\sin\text{x})(1+\sin^2\text{x})}\text{dx}$

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Answer

$\int\frac{2\cos\text{x}}{(1-\sin\text{x})(1+\sin^2\text{x})}\text{dx}$
$\text{put}\sin\text{x}=\text{t}$
$\cos\text{x dx}=\text{dt}$
$\int\frac{2\text{dt}}{(1-\text{t})(1+\text{t}^2)}$
$\frac{2}{(1-\text{t})(1+\text{t}^2)}=\frac{\text{A}}{1-\text{t}}+\frac{\text{Bt}+\text{C}}{1+\text{t}^2}$
$2=\text{A}(1+\text{t}^2)+(\text{Bt}+\text{C})(1-\text{t})$
$\text{put}\ 1-\text{t}=0\ \ |\ \ 2=\text{A}(2)\\\text{t}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \text{A}=1$
Comparing coefficients of t² & t
t² → A + (–B) = 0
B = A
B = 1
t → B – C = 0
B = C = 1
$\int\Big(\frac{1}{1-\text{t}}+\frac{\text{t}+1}{\text{t}^2+1}\Big)\text{dt}$
$\frac{\log(1-\text{t})}{-1}+\int\frac{\text{t}}{\text{t}^2+1}\text{dt}+\int\frac{1}{\text{t}^2+1}\text{dt}$
$-\log(1-\sin\text{x})+\frac{1}{2}\log(\text{t}^2+1)+\tan^{-1}\text{t}+\text{C}$
$-\log(1-\sin\text{x})+\frac{1}{2}\log(\sin^2\text{x}+1)+\tan^{-1}(\sin\text{x})+\text{C}$

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