વિધેયનું સંકલન કરો: $\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}, x \in[0,1]$

(N/A) ધારો કે $I=\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}} d x$.
આપણે જાણીએ છીએ કે $\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}=\frac{\pi}{2}$,તેથી $\cos ^{-1} \sqrt{x}=\frac{\pi}{2}-\sin ^{-1} \sqrt{x}$.
આ કિંમત સંકલનમાં મૂકતા:
$I=\int \frac{\sin ^{-1} \sqrt{x}-(\frac{\pi}{2}-\sin ^{-1} \sqrt{x})}{\frac{\pi}{2}} d x = \frac{2}{\pi} \int (2 \sin ^{-1} \sqrt{x}-\frac{\pi}{2}) d x = \frac{4}{\pi} \int \sin ^{-1} \sqrt{x} d x - x$.
ધારો કે $I_1 = \int \sin ^{-1} \sqrt{x} d x$. $\sqrt{x}=t$ લેતા,$x=t^2$ અને $dx=2t dt$.
$I_1 = \int \sin ^{-1} t \cdot 2t dt = 2 [\frac{t^2}{2} \sin ^{-1} t - \int \frac{t^2}{2\sqrt{1-t^2}} dt] = t^2 \sin ^{-1} t - \int \frac{t^2}{\sqrt{1-t^2}} dt$.
$\int \frac{t^2}{\sqrt{1-t^2}} dt = \int \frac{-(1-t^2)+1}{\sqrt{1-t^2}} dt = -\int \sqrt{1-t^2} dt + \int \frac{1}{\sqrt{1-t^2}} dt = -[\frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin ^{-1} t] + \sin ^{-1} t = -\frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin ^{-1} t$.
તેથી,$I_1 = t^2 \sin ^{-1} t + \frac{t}{2}\sqrt{1-t^2} - \frac{1}{2}\sin ^{-1} t = (t^2 - \frac{1}{2}) \sin ^{-1} t + \frac{t}{2}\sqrt{1-t^2}$.
$t=\sqrt{x}$ પાછા મૂકતા:
$I = \frac{4}{\pi} [(x - \frac{1}{2}) \sin ^{-1} \sqrt{x} + \frac{\sqrt{x}}{2}\sqrt{1-x}] - x + C = \frac{2(2x-1)}{\pi} \sin ^{-1} \sqrt{x} + \frac{2}{\pi} \sqrt{x-x^2} - x + C$.