For \(a> 0\), evaluate \[ \int_0^{\pi/2} \frac{d\theta}{a + \sin^2 \theta} \]
Solution
Set \(t=\tan\theta\). Then \[ \sin^2\theta=\frac{t^2}{1+t^2}, \qquad d\theta=\frac{dt}{1+t^2}. \] Hence \[ \int_0^{\pi/2}\frac{d\theta}{a+\sin^2\theta} =\int_0^\infty\frac{dt}{a+(a+1)t^2}. \] After \(u=t\sqrt{(a+1)/a}\), \[ \int_0^\infty\frac{dt}{a+(a+1)t^2} =\frac1{\sqrt{a(a+1)}}\int_0^\infty\frac{du}{1+u^2}. \] Therefore \[ \boxed{ \int_0^{\pi/2}\frac{d\theta}{a+\sin^2\theta} =\frac{\pi}{2\sqrt{a(a+1)}}.} \]