Question

$\int \frac{si{n}^{-1}\sqrt{x}-co{s}^{-1}\sqrt{x}}{si{n}^{-1}\sqrt{x}+co{s}^{-1}\sqrt{x}}dx.$

Answer 1

$LetI=\int \frac{si{n}^{-1}\sqrt{x}-co{s}^{-1}\sqrt{x}}{si{n}^{-1}\sqrt{x}+co{s}^{-1}\sqrt{x}}dxWeknowthatsi{n}^{-1}\sqrt{x}+co{s}^{-1}\sqrt{x}=\frac{\pi }{2}\Rightarrow co{s}^{-1}\sqrt{x}=\frac{\pi }{2}-si{n}^{-1}\sqrt{x}\therefore I=\int \frac{si{n}^{-1}\sqrt{x}-(\frac{\pi }{2}-si{n}^{-1}\sqrt{x})}{\frac{\pi }{2}}dx=\int \frac{2si{n}^{-1}\sqrt{x}-\frac{\pi }{x}}{\frac{\pi }{2}}dx=\frac{2}{\pi }\int (2si{n}^{-1}\sqrt{x}-\frac{\pi }{2})dx=\frac{4}{\pi }\int si{n}^{-1}\sqrt{x}dx-\int 1dx=\frac{4}{\pi }\int si{n}^{-1}\sqrt{x}dx-x\Rightarrow I=\frac{4}{\pi }{I}_1-x+Cwhere,I_1=\int si{n}^{-1}\sqrt{x}dxput\sqrt{x}=t\Rightarrow x=t^2\Rightarrow dx=2tdt{I}_1=\int si{n}^{-1}t2tdt=2\int \underset{I}{si{n}^{-1}}\underset{II}{t.t}dt=2[si{n}^{-1}t.\frac{t^2}{2}-\int \frac{1}{\sqrt{1-t^2}.}\frac{t^2}{2}dt][UsingintergrationbypartI,\int uvdx=u\int vdx-\int (\frac{d}{dx}u\int vdx)dx]=t^{2}si{n}^{-1}-\int \frac{t^2}{\sqrt{1-t^2}}dt=t^{2}si{n}^{-1}t-\int \frac{-(1-t^2)+1}{\sqrt{1-t^2}}dt=t^{2}si{n}^{-1}t+\int \sqrt{1-t^2}dt-\int \frac{1}{\sqrt{1-t^2}}dt=t^{2}si{n}^{-1}t+\frac{t\sqrt{1-t^2}}{2}+\frac{1}{2}si{n}^{-1}t-si{n}^{-1}t=(t^2-\frac{1}{2})si{n}^{-1}t+\frac{1}{2}t\sqrt{1-t^2}=\frac{1}{2}\left[\right(2x-1)si{n}^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}]=\frac{1}{2}\left[\right(2x-1)si{n}^{-1}\sqrt{x}+\sqrt{x-x^2}]Onputtingthevalueof{I}_{1}inEq.\left(i\right),weget\int \frac{si{n}^{-1}\sqrt{x}-co{s}^{-1}\sqrt{x}}{si{n}^{-1}\sqrt{x}+co{s}^{-1}\sqrt{x}}dx=\frac{2}{\pi }\left[\right(2x-1)si{n}^{-1}\sqrt{x}+\sqrt{x-x^2}]-x+C$