Wolfram Alpha says that this integral does not converge
$$\int^{\pi/2}_{-\pi/2}\frac{\mathrm{d}x}{a+b\sin(x)},$$ $a,b \in \mathbb{R}^+$ with also the condition $b \geq a$.
Is it true? Making a fast calculation using the residue theorem I have found that it is equal to zero.
I have used the fact that $$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$ and the substitution $z = e^{ix}$.
On
Robert Israel provided the good explanations.
Concerning the antiderivative, using the tangent half-angle substitution, you should arrive to $$I=\int \frac{dx}{a+b\sin(x)}=\int\frac{dt}{at^2+2bt+a}$$ If $b \neq a$, then $$I=\frac{2 \tan ^{-1}\left(\frac{a \tan \left(\frac{x}{2}\right)+b}{\sqrt{a^2-b^2}}\right)}{\sqrt{a^2-b^2}}$$ from which (if $a>b$)$$J=\int^{\pi/2}_{-\pi/2}\frac{dx}{a+b\sin(x)}=\frac{2 \left(\tan ^{-1}\left(\frac{a-b}{\sqrt{a^2-b^2}}\right)+\tan ^{-1}\left(\frac{a+b}{\sqrt{a^2-b^2}}\right)\right)}{\sqrt{a^2-b^2}}$$ Now, using $$\tan^{-1}(p)+\tan^{-1}(q)=\tan ^{-1}\left(\frac{p+q}{1-p\, q}\right)$$ should give (since, here, $p\,q=1$) $$J=\frac \pi{\sqrt{a^2-b^2}} $$
If $b \ge a > 0$ there is a singularity at $x = \arcsin(-a/b)$, and the integral diverges. However, if $b > a > 0$ it will have a Cauchy principal value.