Question

Integrate :

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

Answer 1

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

$=\int \frac{6{\cos }^{2}X+2\cos X-23}{(t-1{)}^2(4+t{)}^2}X-dt\left(\sin X\right)$

i.e, $\cos x=t\Rightarrow -\sin xdx=dt$ {substitution}

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

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

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

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

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

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

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

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

$=-\frac{5t}{2}{\tan }^{-1}\left(\frac{t}{2}\right)+C$