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

Hint:
By inspection,
$$\int\frac{\sin x\cos x}{(3+4\cos x)^3}dx=-\int \frac{t}{(3+4t)^3}dt$$
which can be decomposed with
$$4t=(3+4t)-3,$$ to yield elementary functions.
