<s>[INST] Evaluating $ \int \frac{1}{5 + 3 \sin(x)} ~ \mathrm{d}{x} $. 
What is the integral of: $\int \frac{1}{5+3\sin x}dx$

My attempt:
Using: $\tan \frac x 2=t$, $\sin x = \frac {2t}{1+t^2}$, $dx=\frac {2dt}{1+t^2}$ we have:
$\int \frac{1}{5+3\sin x}dx= 2\int \frac 1 {5t^2+6t+5}dt $
I'll expand the denominator: $5t^2+6t+5=5((t+\frac 3 5 )^2+1-\frac 1 4 \cdot (\frac 6 5)^2)=5((t+\frac 3 5)^2+0.64)$. So:
$2\int \frac 1 {5t^2+6t+5}dt = \frac 2 5 \int \frac 1{(t+\frac 3 5)^2+0.64}dt=\frac 2 5(\frac 5 4\arctan((t+\frac 3 5)\frac 5 4))=\frac 1 2 \arctan(\frac{5t+3}{4}) $
But if I'll place $\tan \frac x 2=t$ I won't be able to simplify it further and since the online calculator's answers don't have $\tan \frac x 2$ there, I believe I made a mistake. What is wrong with what I did and is there a better way to do it?
 [/INST] Taking from 
$5t^2$+6t+5, instead of expanding, complete the square.
=5(t+$\frac{6}{10})^2$-$\frac{16}{5}$
Rewrite in the form
=$\frac{-1}{5}$$\int$$\frac{1}{16/25-(t+6/10)^2}$
=$\frac{-1}{5}$[$\frac{5}{4}$$tanh^{-1}$($\frac{t+6/10}{4/5})]$+C
=$\frac{-1}{4}$[$tanh^{-1}$($\frac{5t}{4}$+$\frac{3}{2}$)]+C
Still getting answer in terms of tan(x/2).
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