Question
Integrate the functions in Exercises:
$\frac{\sin\text{x}}{(1+\cos\text{x})^2}$

Answer

$ \text{Let I}=\int\frac{\sin\text{x}}{(1 +\cos\text{x})^2}\text{ dx}=-\int\frac{-\sin\text{x}}{(1+\cos\text{x})^2}\text{ dx} \ \ \ \ \ \ \ \ ...\text{(i)} $
Putting $1+\cos\text{x}=\text{t}\ \ \ \Rightarrow\ \ \ -\sin\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow\ \ \ -\sin\text{x dx}=\text{dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=-\int\frac{\text{dt}}{\text{t}^2}=-\int\text{t}^{-2}\text{ dt}=\frac{-\text{t}^{-1}}{-1}+\text{c}$
$=\frac{1}{1+\cos\text{x}}+\text{c}$

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