Evaluation of $\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$ Evaluation of $\displaystyle \int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$
$\bf{My\; Solution::}$ Given $\displaystyle \int\frac{1}{\sin^2 x\cdot (5+4\cos x)}dx = \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx$
Now Using Partial fraction for $\displaystyle \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}$
Now Let $\cos x= y\;,$ and Let $\displaystyle \frac{1}{(1-y)(1+y)(5+4y)} = \frac{A}{1+y}+\frac{B}{1-y}+\frac{C}{5+4y}$
after solving We Get $\displaystyle A = \frac{1}{2}$ and $\displaystyle B = -\frac{1}{18}$ and $\displaystyle C = -\frac{16}{9}$
So $\displaystyle \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx = \frac{1}{2}\int \frac{1}{1+\cos x}dx - \frac{1}{18}\int\frac{1}{1-\cos x}dx - \frac{16}{9}\int \frac{1}{5+4\cos x}dx$
And after that we can solve easily Like for
$\displaystyle \int\frac{1}{1+\cos x}dx = \int\frac{1-\cos x}{\sin^2 x}dx = \int \left(\csc^2 x-\csc x\cdot \cot x\right)dx = -\cot x +\csc x+\mathcal{C}$
My Question is , Is there is any other method by which we can solbe the above question.
OR without using partial fraction,
Thanks

Doing the same as Aditya (Weierstrass substitution), you arrive to $$\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx=\int \frac{\left(t^2+1\right)^2}{2 t^2 \left(t^2+9\right)}dt$$ Now, using partial fraction decomposition $$\frac{\left(t^2+1\right)^2}{2 t^2 \left(t^2+9\right)}=\frac{1}{18 t^2}-\frac{32}{9 \left(t^2+9\right)}+\frac{1}{2}$$ and integration is not so difficult, leading to $$\int \frac{\left(t^2+1\right)^2}{2 t^2 \left(t^2+9\right)}dt=\frac{1}{54} \left(27 t-\frac{3}{t}-64 \tan ^{-1}\left(\frac{t}{3}\right)\right)$$ or, going back to $x$ $$\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx=\frac{1}{2} \tan \left(\frac{x}{2}\right)-\frac{1}{18} \cot
   \left(\frac{x}{2}\right)+\frac{32}{27} \tan ^{-1}\left(3 \cot
   \left(\frac{x}{2}\right)\right)$$
Just as Aditya said : OOps PF!
But, again, this is almost what you did well.
