Mathematics

Evaluate the following integral:
$$\int { \cfrac { \sin { x }  }{ { \left( 1+\cos { x }  \right)  }^{ 2 } }  } dx$$


SOLUTION
Let
$$t=1+\cos{x}\Rightarrow\,dt=-\sin{x}\,dx$$

$$\Rightarrow\,-dt=\sin{x}\,dx$$

$$\displaystyle\int{\dfrac{\sin{x}dx}{{\left(1+\cos{x}\right)}^{2}}}$$

$$=-\displaystyle\int{\dfrac{dt}{{t}^{2}}}$$

$$=-\displaystyle\int{{t}^{-2}\,dt}$$

$$=-\dfrac{{t}^{-2+1}}{-2+1}+c$$

$$=\dfrac{1}{t}+c$$

$$=\dfrac{1}{1+\cos{x}}+c$$ where $$t=1+\cos{x}$$
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Subjective Medium Published on 17th 09, 2020
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