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

Yes there exists another one(OOps PF!):
$$\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx
\stackrel{t=\tan x/2}=\int\frac{t^6+3 t^4+3 t^2+1}{4 t^4+36 t^2}dx\\
= \int(t^2/4+128/(9 (t^2+9))+1/(36 t^2)-3/2)dx=...$$
Similiar to your method. Your's is best, why are you looking for other methods?
