<s>[INST] Evaluation of $\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$ 
Evaluation of $$\int\frac{\sin 2x}{(3+4\cos  x)^3}\,dx$$

$\bf{My\; Try::}$ Let $$\displaystyle \int\frac{\sin 2x}{(3+4\cos  x)^3}dx = 2\int\frac{\sin x\cos x}{(3+4\cos x)^3}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^3 x.$
So we get $$\displaystyle I = 2\int\frac{\sec x\cdot \tan x}{(3\sec x+4)^3}dx$$
Now Put $\displaystyle (3\sec x+4) = t\;,$ Then $3\sec x\cdot \tan xdx = dt$
So we get $$\displaystyle I = \frac{2}{3}\int \frac{1}{t^3}dt = -\frac{1}{3t^2}+\mathcal{C} = -\frac{1}{3(3\sec x+4)^2}+\mathcal{C}$$
But  answer is http://www.wolframalpha.com/input/?i=INTEGRATION+OF+%28sin+2x%29%2F%283%2B4cos+x%29%5E3
Where I gave Done Wrong,
Thanks
 [/INST] Going by your work so far we have
$$
2\int \frac{\sin \theta \cos \theta}{(3+4\cos \theta)^3}d\theta
$$
Lets set $u = 3+4\cos \theta\to du = -4\sin\theta d\theta$
thus your integral becomes
$$
-\frac{2}{16}\int \frac{u-3}{u^3}du=?
$$
This yields the result you desire! Though the interesting question is how to go from your expression
$$
\begin{align}
-\frac{1}{3}\frac{1}{(3\sec x + 4)^2} + C_1 + C_2 &=& -\frac{1}{3}\left[\frac{\cos^2 x}{(3+4\cos x)^2}-3C_1\right]+ C_2\\
&=&-\frac{1}{3}\left[\frac{\cos^2 x-3C_1(3+4\cos x)^2}{(3+4\cos x)^2}\right]+ C_2\\
&=&-\frac{1}{3}\left[\frac{\cos^2 x-3C_1\left(9+24\cos x + 16\cos^2 x\right)}{(3+4\cos x)^2}\right]+ C_2\\
&=&-\frac{1}{3}\left[\frac{\left(1-3\cdot 16\cdot C_1\right)\cos^2 x-3\cdot 9\cdot C_1-3\cdot 24\cdot C_1\cos x }{(3+4\cos x)^2}\right]+ C_2
\end{align}
$$
To be similar to Wolfram you need to set $\cos^2 x$ coefficient to zero so we have $1-3\cdot 16\cdot C_1 = 0\implies C_1 = \frac{1}{3\cdot 16}$ so we obtain
$$
-\frac{1}{3}\frac{1}{(3\sec x + 4)^2} + C_1 + C_2 = -\frac{1}{3}\left[\frac{- \frac{9}{16}-\frac{24}{16}\cos x }{(3+4\cos x)^2}\right]+ C_2
$$
this becomes
$$
\left[\frac{\frac{3}{16}+\frac{8}{16}\cos x }{(3+4\cos x)^2}\right]+ C_2
$$
which finally becomes
$$
\frac{1}{16}\frac{3+8\cos x }{(3+4\cos x)^2}+ C_2
$$
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