Evaluate the following integrals: = (1+ )^2 dx - Vidyadip

Question

Evaluate the following integrals:
$=\int\frac{\sin\text{x}}{(1+\cos\text{x})^2}\text{dx}$

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Answer

Let I $=\int\frac{\sin\text{x}}{(1+\cos\text{x})^2}\text{dx}\ ....(1)$
Let $1+\cos\text{x}=\text{t}$ then,
$\text{d}(1+\cos\text{x})=\text{dt}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\Rightarrow\sin\text{x dx}=-\text{dt}$
Putting $1+\cos\text{x}=\text{t}$ and $\sin\text{dx}=-\text{dt}$ in equation (2), we get
$\text{I}=\int\frac{-\text{dt}}{\text{t}^2}$
$=-\int\text{t}^{-2}\text{dt}$
$=-\big(-1\text{t}^{-1}\big)+\text{C}$
$=\frac{1}{\text{t}}+\text{C}$
$=\frac{1}{1+\cos\text{x}}+\text{C}$
$\therefore\text{I}=\frac{1}{1+\cos\text{x}}+\text{C}$

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