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?
HINT:
Use Weierstrass substitution, $$\tan\frac x2=t$$
$$\implies\sin x=\frac{2t}{1+t^2}$$ and $$\frac x2=\arctan t\implies dx=\frac{2\ dt}{1+t^2}$$
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.
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))
You may refer to this link to answer your question. It is almost exactly similar with your question. I hope this helps.
