<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] Hint:
From what you got on that calculator, use the following identities to simplify it:


*

*$\sin(x)=2\sin(x/2)\cos(x/2)$

*$1+\cos(x)=2\cos^2(x/2)$


You'll see that the results are same.
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