Question:

Put $x = a\cos^2\theta + b\sin^2\theta$. The given integral becomes<br>$$I = \int \frac{2(b-a)\sin\theta\cos\theta d\theta}{\{(a\cos^2\theta + b\sin^2\theta - a)(b - a\cos^2\theta - b\sin^2\theta)\}^{\frac{1}{2}}}$$<br>$$= \int \frac{2(b-a)\sin\theta\cos\theta d\theta}{(b-a)\sin\theta\cos\theta} = \left(\frac{b-a}{b-a}\right)\int 2d\theta = 2\theta + C = 2\sin^{-1}\sqrt{\frac{x-a}{b-a}} + C$$