Question

Find the value of \[ \int_0^{\pi/2} \frac{x}{a^2 \sin^2 x + b^2 \cos^2 x} \, dx. \]

Hint:

When solving integrals with trigonometric terms in the denominator, symmetry and standard results can simplify the process.

Answer 1

Lettheintegralbe: \[ I=\int_0^{\pi/2}\frac{x}{a^2\sin^2x+b^2\cos^2x}\,dx. \] Usingthepropertyofdefiniteintegrals: \[ \int_0^af(x)\,dx=\int_0^af(a-x)\,dx, \] wesubstitute\(x\to\pi/2-x\).Thisgives: \[ I=\int_0^{\pi/2}\frac{\pi/2-x}{a^2\cos^2x+b^2\sin^2x}\,dx. \] Addingthesetwoformsof\(I\),weget: \[ 2I=\int_0^{\pi/2}\frac{x}{a^2\sin^2x+b^2\cos^2x}\,dx+\int_0^{\pi/2}\frac{\pi/2-x}{a^2\cos^2x+b^2\sin^2x}\,dx. \] Simplify: \[ 2I=\frac{\pi}{2}\int_0^{\pi/2}\frac{1}{a^2\sin^2x+b^2\cos^2x}\,dx. \] Now,let: \[ J=\int_0^{\pi/2}\frac{1}{a^2\sin^2x+b^2\cos^2x}\,dx. \] Usingthestandardresultforthisintegral: \[ J=\frac{\pi}{2ab}. \] Thus: \[ 2I=\frac{\pi}{2}\cdot\frac{\pi}{2ab}\quad\Rightarrow\quadI=\frac{\pi^2}{4ab}. \]