$\displaystyle \displaystyle { \int \frac{1}{2-\sin x} dx } $
any hints how I can integrate it
Sub. $\displaystyle x = \frac{\pi}{2} - 2t $
we have
$\displaystyle \int \frac{dx}{2 - \sin(x) } = - 2 \int \frac{dt}{ 2 - \cos(2t) } $
$\displaystyle = -2 \int \frac{dt}{ 1 + (1 - \cos(2t) )} $
$\displaystyle = -2 \int \frac{dt}{1 + 2\sin^2(t) } $
$\displaystyle = -2 \int \frac{dt}{ \cos^2(t) + 3\sin^2(t) } $
$\displaystyle = -2 \int \frac{ \sec^2(t) ~dt }{ 1 + 3 \tan^2(t) } $
$\displaystyle = - \frac{2}{\sqrt{3} } \tan^{-1}[ \sqrt{3} \tan(t) ] + C $
Then back substitution .