Calculation

(1). $\dps{\int_a^b \frac{\rd x}{\sqrt{(x-a) (b-x)}}\ (b>a)}$.

(2). $\dps{\int_{-1}^1 \frac{\rd x}{(a-x) \sqrt{1-x^2}}\ (a>1)}.$

Answer:

(1). $$\beex \bea \mbox{original points}&=\int_{-\frac{b-a}{2}}^\frac{b-a}{2} \frac{\rd t}{\sqrt{\sex{t+\frac{b-a}{2}}\sex{ \frac{b-a}{2}-t}}}\\ &=\int_{-\frac{b-a}{2}}^{\frac{b-a}{2}} \frac{\rd t}{\sqrt{\sex{\frac{b-a}{2}}^2-t^2}}\\ &=\INT_{-\FRAC{\PI}{2}}^\FRAC{\PI}{2} \frac{1}{\frac{b-a}{2}\cos \tt}\cdot \frac{b-a}{2}\cos \tt\rd \tt\\ &= \pi. \eea \eeex$$

(2). $$\beex \bea \mbox{original points}&=\int_{-\frac{\pi}{2}}^\frac{\pi}{2} \frac{\cos t\rd t}{(A-\sin t) \cos t}\quad\sex{x= \sin t}\\ &=\int_{-\frac{\pi}{2}}^\frac{\pi}{2} \frac{\rd t}{a-\sin t}\\ &=\int_{-\frac{\pi}{2}}^0+\int_0^\ FRAC{\PI}{2} \frac{\rd t}{a-\sin t}\\ &=\int_0^\frac{\pi}{2}\frac{\rd y}{a+\sin y}+\int_0^\frac{\pi}{2} \frac{\rd T}{a-\sin t}\quad\sex{t=-y}\\ &=\int_0^\frac{\pi}{2} \frac{2a}{a^2-\sin^2t}\rd t\\ &=2a\int_0^\frac{\pi}{2} \frac{\sin^2t+\cos^2t}{(a^2-1) \sin^2t+a^2\cos^2t}\rd t\\ &=2a\int_0^\frac{\pi}{2} \frac{\rd \tan t}{a^2+ (a^2-1) \tan^2t}\\ &=\frac{2}{\sqrt{a^2-1}} \int_0^\frac{\pi}{2} \frac{\rd \sex{\frac{\sqrt{a^2-1}}{a} \tan t}}{1+\sex{\ Frac{\sqrt{a^2-1}}{a} \tan t}^2}\\ &=\frac{2}{\sqrt{a^2-1}}\cdot \frac{\pi}{2}\\ &=\frac{\pi}{\sqrt{a^2-1}} . \eea \eeex$$

[Typical problems and methods in mathematical analysis of Periven exercises reference Solutions]4.5.1