# 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

• Once again I said, you have asked so many good questions and each question has some good answers but the problem is you almost never upvote nor accept the answer of your own questions. I don't think this is a good manner in M.SE. – Tunk-Fey Sep 15 '14 at 18:51
• Sorry Tunk-Fey, But I have always upvoted answer. and may be i have not accept any answer because it is not the answer which i have asked.(althought i have accept answer).or may be i have missed it..any way next time i will always accept answers, Thanks – juantheron Sep 15 '14 at 19:48

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?

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.