Evaluate the integral $\int\frac{1}{2+3\sin x}\,\text{d}x$. Please evaluate the integral,
$$\int \frac{1}{2+3\sin x}\,\text{d}x.$$
What I have tried is to substitute $\sin x = \sqrt{1-x^x}$ but I was stuck in a maze. Also, I did look a the wolfram solution. Can anyone propose a different solution from Wolfram, perhaps simpler with a bit of explanation?
 A: HINT:
Use Weierstrass substitution, $$\tan\frac x2=t$$
$$\implies\sin x=\frac{2t}{1+t^2}$$  and $$\frac x2=\arctan t\implies dx=\frac{2\ dt}{1+t^2}$$
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A: use Weierstrass substitution: $\sin\left(x\right)=\frac{2t}{1+t^2}$, where $t=\tan\left(\frac{x}{2}\right)$
$⇔2\arctan\left(t\right)=x$,$dx=\frac{2dt}{1+t^2}$
$$\int_{ }^{ }\frac{dx}{2+3\sin\left(x\right)}
$$$$=\int_{ }^{ }\frac{\frac{2dt}{1+t^2}}{2+3\left(\frac{2t}{1+t^2}\right)}
$$$$=\int_{ }^{ }\frac{2dt}{2t^2+6t+2}
$$$$=\int_{ }^{ }\frac{dt}{t^2+3t+1}$$
$$=\int_{ }^{ }\frac{dt}{\left(t+\frac{3}{2}\right)^2-\frac{5}{4}}$$
substitute $t+\frac{3}{2}=u$
$$\int_{ }^{ }\frac{du}{u^2-\frac{5}{4}}$$
now we need another substitution:$\frac{2}{\sqrt{5}}u=v,\frac{dv}{du}=\frac{2}{\sqrt{5}}⇔du=\frac{\sqrt{5}}{2}dv$
$$\frac{4}{5}\int_{ }^{ }\frac{\frac{\sqrt{5}}{2}dv}{v^2-1}$$
$$\frac{2}{\sqrt{5}}\arctan\left(v\right)+C$$
undo substitution $\frac{2}{\sqrt{5}}u=v$, we have:$$\frac{2}{\sqrt{5}}\arctan\left(\frac{2}{\sqrt{5}}u\right)+C$$
undo substitution $t+\frac{3}{2}=u$, we have:$$\frac{2}{\sqrt{5}}\arctan\left(\ \frac{2\tan\left(\frac{x}{2}\right)+3}{\sqrt{5}}\right)+C$$
we're done.
A: write sin x as 2*sin (x/2)*cos(x/2) and divide numerator and denominator by cos^2(x/2). Then substitute tan(x/2) for t,complete perfect square in the denominator(there will be a constant as well).final answer will be (2*tan inverse(2*tan (x/2)+3))
