Mathematics

Evaluate the following integral:

$$\displaystyle \int { \cfrac { 2\cos { x }  }{ (1-\sin { x } )(1+\sin ^{ 2 }{ x } ) }  } dx$$


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

Let $$t=\sin{x}\Rightarrow\,dt=\cos{x}\,dx$$

$$=2\displaystyle\int{\dfrac{dt}{\left(1-t\right)\left(1+{t}^{2}\right)}}$$
Solving by the method of partial fractions,

$$\dfrac{1}{\left(1-t\right)\left(1+{t}^{2}\right)}=\dfrac{A}{\left(1-t\right)}+\dfrac{Bt+C}{\left(1+{t}^{2}\right)}$$

$$\Rightarrow\,1=A\left(1+{t}^{2}\right)+\left(Bt+C\right)\left(1-t\right)$$
Put $$t=1\Rightarrow\,1=2A$$

$$\Rightarrow\,A=\dfrac{1}{2}$$
Put $$t=0\Rightarrow\,1=A+C$$

$$\Rightarrow\,C=1-A=1-\dfrac{1}{2}=\dfrac{1}{2}$$
Put $$t=-1\Rightarrow\,1=2A-2B+2C$$

$$\Rightarrow\,2\left(A+C\right)-2B=1$$
$$\Rightarrow\,2\times\,1-2B=1$$ since $$A+C=1$$

$$\Rightarrow\,-2B=1-2=-1$$
$$\Rightarrow\,B=\dfrac{1}{2}$$

$$\therefore\,A=\dfrac{1}{2},\,B=\dfrac{1}{2}$$ and $$C=\dfrac{1}{2}$$
$$\dfrac{1}{\left(1-t\right)\left(1+{t}^{2}\right)}=\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{\left(1-t\right)}}+\dfrac{1}{2}\displaystyle\int{\dfrac{\left(t+1\right)dt}{\left(1+{t}^{2}\right)}}$$

$$=\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{\left(1-t\right)}}+\dfrac{1}{4}\displaystyle\int{\dfrac{2t\,dt}{\left(1+{t}^{2}\right)}}+\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{\left(1+{t}^{2}\right)}}$$

$$=-\dfrac{1}{2}\log{\left|1-t\right|}+\dfrac{1}{4}\log{\left|1+{t}^{2}\right|}+\dfrac{1}{2}{\tan}^{-1}{t}+c$$

$$=-\dfrac{1}{2}\log{\left|1-\sin{x}\right|}+\dfrac{1}{4}\log{\left|1+{\sin}^{2}{x}\right|}+\dfrac{1}{2}{\tan}^{-1}{\left(\sin{x}\right)}+c$$ where $$t=\sin{x}$$

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