Evaluating the indefinite trigonometric integral $\int \frac{dx}{(3+2\sin x)^2}$ 
$$\int \frac{dx}{(3+2 \sin x)^2}$$

ATTEMPT:-
Re Writing the integral as:
$I=\int \frac{2\cos x \sec x \,dx}{2(3+2 \sin x)^2}$ and using by parts:-
$\int u\,dv= uv-\int v\,du$
Here $u=\sec x \implies du=\sec x \tan x \,dx$
$\quad$ $dv=\frac{2\cos x \,dx}{2(3+2sinx)^2} \implies v=\frac{-1}{2(3+2\sin x)}$
$I=\frac{-\sec x\,dx}{2(3+2\sin x)} +\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$
Let $I'=\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$
$\implies I'=\int \frac{\sin x \,dx}{2\cos^2x(3+2\sin x)}$
$\implies I'=\int \frac{\sin \,dx}{2(1-\sin^2x)(3+2\sin x)}$ 
$\implies I'=\int \frac{-dx}{4(1+\sin x)} +\frac{3dx}{5(2\sin x+3)} + \frac{-dx}{20(\sin x-1)}$ which can be easily done by weierstrass's substitution.
But I am not able to modify my answer as given in the text.
Text Ans:-$\frac{2\cos x\,dx}{5(3+2\sin x)^2} + \frac{2}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}} +c.$
My Ans:-$\frac{-\sec x\,dx}{2(3+2\sin x)} +  \frac{6}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}}-10\tan x+15\sec x-15.$

Let $\displaystyle I = \int\frac{1}{(3+2\sin x)^2}dx\;,$ Now Put $\displaystyle \frac{2+3\sin x}{(3+2\sin x)} = t\;,$ 
Then $$\displaystyle \frac{(3+2\sin x)\cdot 3\cos x-(2+3\sin x)\cdot 2\cos x}{(3+2\sin x)^2}dx = dt$$
so we get $$\frac{5\cos x}{(3+2\sin x)^2}dx = dt\Rightarrow \frac{1}{(3+2\sin x)^2}dx = \frac{1}{5\cos x}dt.$$
So We get $$I = \frac{1}{5}\int \frac{1}{\cos x}dt$$
Now Above we have $$\frac{2+3\sin x}{3+2\sin x}=t\Rightarrow \sin x = \frac{2-3t}{2t-3}.$$
So we get $$\cos x= \sqrt{1-\sin^2 x} = \frac{\sqrt{5}\cdot \sqrt{1-t^2}}{2t-3}.$$
So we get $$I = \frac{1}{5\sqrt{5}}\int\frac{(2t-3)}{\sqrt{1-t^2}}dt = \frac{2}{5\sqrt{5}}\int\frac{t}{\sqrt{1-t^2}}dt-\frac{3}{5\sqrt{5}}\int\frac{1}{\sqrt{1-t^2}}dt$$
