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?
Best Answer
HINT:
Use Weierstrass substitution, $$\tan\frac x2=t$$
$$\implies\sin x=\frac{2t}{1+t^2}$$ and $$\frac x2=\arctan t\implies dx=\frac{2\ dt}{1+t^2}$$