Mathematics

Integrate :
$$\int {\cfrac{{6\sin \left( x \right)co{s^2}\left( x \right) + sin\left( {2x} \right) - 23\sin \left( x \right)}}{{{{\left( {\cos \left( x \right) - 1} \right)}^2}\left( {5 - {{\sin }^2}\left( x \right)} \right)}}dx} $$


SOLUTION
$$\displaystyle =\int\dfrac{6\sin X \cos^2 X+2\sin X\cos X-23\sin X}{(\cos X-1)^2(4+\cos^2X)}$$

$$\displaystyle =\int\dfrac{6\cos^2 X+2\cos X-23}{(t-1)^2(4+t)^2}X-dt(\sin X)$$
i.e, $$\cos\,x=t\Rightarrow -\sin\,xdx=dt$$ {substitution}

$$\displaystyle \int-\dfrac{-(6t^2+2t-23)dt}{(t-1)^2(4+t^2)}dt$$

$$\displaystyle \int\dfrac{23-2t-6t^2}{(t-1)^2(4+t)^2}$$

$$\Rightarrow \boxed{\dfrac{23-2t-6t^2}{(t-1)^2(4+t)^2}=\dfrac{At+B}{4+t^2}+\dfrac{C}{t-1}+\dfrac{D}{(t-1)^2}}$$

By solving we get $$D=3;A=4;B=-5;C=-4$$

$$\displaystyle \int\Rightarrow \dfrac{4t-5dt}{4+t^2}+\int\dfrac{-4}{t-1}dt+\dfrac{3}{(t-1)^2}dt$$

$$\displaystyle \int\dfrac{4t}{4+t^2}dt-\int\dfrac{5}{4+t^2}dt-4\log(t-1)$$

$$=2\log(4+t^2)-4\log(t-1)+\dfrac{3}{t-1}+C$$

$$=2\log(4+t^2)-4\log(t-1)+\dfrac{3}{t-1}$$

$$=-\dfrac{5t}{2}\tan^{-1}\left(\dfrac{t}{2}\right)+C$$
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Subjective Medium Published on 17th 09, 2020
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