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Question Number 159823 by tounghoungko last updated on 21/Nov/21

Answered by Ar Brandon last updated on 21/Nov/21

$$I={\int }_0^1\frac{\sqrt{1-x}}{\sqrt{1+x}}{\sin }^{-1}xdx={\int }_0^1\frac{1-x}{\sqrt{1-x^2}}{\sin }^{-1}xdx$$$$={\int }_0^1\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx-{\int }_0^1\frac{x}{\sqrt{1-x^2}}{\sin }^{-1}xdx$$$$={\left[\frac{{\left({\sin }^{-1}x\right)}^2}{2}\right]}_0^1-{[-\sqrt{1-x^2}{\sin }^{-1}x+\int \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}dx]}_0^1$$$$=\frac{{\pi }^2}{8}-{\int }_0^{1}dx=\frac{{\pi }^2}{8}-1$$

Commented by Ar Brandon last updated on 21/Nov/21

$${\int }_0^1\frac{{\sin }^{-1}\left(x\right)}{\sqrt{1-x^2}}dx.\mathrm{Let}s={\sin }^{-1}x\Rightarrow ds=\frac{dx}{\sqrt{1-x^2}}$$$${\int }_0^1\frac{x}{\sqrt{1-x^2}}{\sin }^{-1}xdx$$$$\mathrm{By}\mathrm{part}\{\begin{array}{l}u\left(x\right)={\sin }^{-1}x\\ {\mathrm{v}}^{\prime }\left(x\right)=\frac{x}{\sqrt{1-x^2}}\end{array}$$

Commented by tounghoungko last updated on 21/Nov/21

$$awesome$$

Answered by Kunal12588 last updated on 21/Nov/21

$${\int }_0^1\frac{\sqrt{1-x}{\sin }^{-1}x}{\sqrt{1+x}}dx$$$$I=\int \frac{\sqrt{1-x}{\sin }^{-1}x}{\sqrt{1+x}}dx$$$$u={\sin }^{-1}x,dv=\sqrt{\frac{1-x}{1+x}}dx$$$$\Rightarrow du=\frac{1}{\sqrt{1-x^2}}dx$$$$v=\int \sqrt{\frac{1-x}{1+x}}dx[takingx=\sin u]$$$$v=\int \sqrt{\frac{1-\sin u}{1+\sin u}}\times \cos udu=\int \frac{1-\sin u}{\cos u}\times \cos udu$$$$\Rightarrow v=u+\cos u={\sin }^{-1}x+\sqrt{1-x^2}$$$$I={\sin }^{-1}x({\sin }^{-1}x+\sqrt{1-x^2})-\int \frac{{\sin }^{-1}x+\sqrt{1-x^2}}{\sqrt{1-x^2}}dx$$$$I={\sin }^{-1}x({\sin }^{-1}x+\sqrt{1-x^2})-\int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx-\int dx$$$$lett={\sin }^{-1}x\Rightarrow dt=\frac{1}{\sqrt{1-x^2}}dx$$$$I={\sin }^{-1}x({\sin }^{-1}x+\sqrt{1-x^2})-\int tdt-x$$$$I={\sin }^{-1}x({\sin }^{-1}x+\sqrt{1-x^2})-x-\frac{1}{2}{t}^2$$$$I={\sin }^{-1}x({\sin }^{-1}x+\sqrt{1-x^2})-x-\frac{1}{2}{\left({\sin }^{-1}x\right)}^2+C$$$$\Rightarrow I=\frac{1}{2}{\left({\sin }^{-1}x\right)}^2+\sqrt{1-x^2}{\sin }^{-1}x-x+C$$$${\int }_0^1\frac{\sqrt{1-x}{\sin }^{-1}x}{\sqrt{1+x}}dx$$$$=[\frac{1}{2}{\left(\frac{\pi }{2}\right)}^2+0\times \frac{\pi }{2}-1]-[\frac{1}{2}\times 0+0-0]$$$${\int }_0^1\frac{\sqrt{1-x}{\sin }^{-1}x}{\sqrt{1+x}}dx=\frac{{\pi }^2}{8}-1$$

Commented by tounghoungko last updated on 21/Nov/21

$$yess$$

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