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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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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