Evaluate the integral $\int\frac{1}{2+3\sin x}\,\text{d}x$.

Question

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?

Answer 1

HINT:

Use Weierstrass substitution, $$\tan\frac x2=t$$

$$\implies\sin x=\frac{2t}{1+t^2}$$ and $$\frac x2=\arctan t\implies dx=\frac{2\ dt}{1+t^2}$$

Answer 2

You may refer to this link to answer your question. It is almost exactly similar with your question. I hope this helps.

Answer 3

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))