Calculation of $\int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$ Calculation of $\displaystyle \int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$
$\bf{My\; Try}::$ Using $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
Let $\displaystyle I = \int_{0}^{\pi}\frac{1}{\left(5+\frac{4-4\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}\right)^2}dx = \int_{0}^{\pi}\frac{1+\tan^2 \frac{x}{2}}{\left(9+\tan^2 \frac{x}{2}\right)^2}dx$
$\displaystyle I = \int_{0}^{\pi}\frac{\sec^2 \frac{x}{2}}{\left(9+\tan^2 \frac{x}{2}\right)^2}dx$
Now Let $\tan \frac{x}{2} = t$ and $\sec^2 \frac{x}{2}dx = 2dt$
$\displaystyle I = 2\int_{0}^{\infty}\frac{1}{(9+t^2)^2dt}$
Now I did not understand how can i solve after that
Help Required
Thanks

HINT
Let $$u = \tan(\frac{x}{2})$$
Using the trig. identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$ (1)
$$u = \tan(\frac{x}{2}) = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$$
Squaring both sides $$u^2 \cdot \cos^2(\frac{x}{2}) = \sin^2(\frac{x}{2})$$
Using the trig. identity $\sin^2(x) + cos^2(x) = 1$ (2) This becomes
$$u^2 \cdot \cos^2(\frac{x}{2}) = 1 - \cos^2(\frac{x}{2})$$
$$u^2 \cdot \cos^2(\frac{x}{2}) + \cos^2(\frac{x}{2}) = 1 \implies \cos^2(\frac{x}{2}) \cdot (u^2 + 1) = 1 \implies \cos^2(\frac{x}{2}) = \frac{1}{(u^2 + 1)}$$
Using trig identity (2) again becomes
$$\sin^2(\frac{x}{2}) = 1 - \frac{1}{(u^2 + 1)} = \frac{u^2}{(u^2 + 1)}$$
Then 
$$\cos(x) = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = \frac{1}{(u^2 + 1)} - \frac{u^2}{(u^2 + 1)} = \frac{1 - u^2}{u^2 + 1}$$
Back to the original statement, $$u = \tan(\frac{x}{2})$$ inverse $\tan$ both sides to get
$$\tan^{-1}u = \frac{x}{2}$$
Then differentiate both sides and using trig. identity $\frac{d}{dx} \tan^{-1}(x) = \frac{1}{x^2 + 1}$
$$dx = \frac{2 \cdot du}{u^2 + 1}$$
Substitute these back into the integral and solve. 
