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Question Number 108095 by bemath last updated on 14/Aug/20

$$\frac{\mathrm{\infty }\mathcal{B}e\mathcal{M}ath\mathrm{\infty }}{\mathrm{♠}}$$$$\int 2x{\cos }^{-1}\left(x\right)dx$$

Answered by bemath last updated on 14/Aug/20

Answered by Dwaipayan Shikari last updated on 14/Aug/20

$$x^2{cos}^{-1}x+\int \frac{x^2}{\sqrt{1-x^2}}dx$$$$x^2{cos}^{-1}x-\int \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}$$$$x^2{cos}^{-1}x-\int \sqrt{1-x^2}+{sin}^{-1}x$$$$x^2{cos}^{-1}x+{sin}^{-1}x-\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}{sin}^{-1}x+C$$$$x^2{cos}^{-1}x+\frac{1}{2}{sin}^{-1}x-\frac{x}{2}\sqrt{1-x^2}+C$$

Answered by mathmax by abdo last updated on 14/Aug/20

$$\mathrm{I}=2\int \mathrm{x}\mathrm{arcosx}\mathrm{dx}\mathrm{by}\mathrm{parts}$$$$\mathrm{I}=2\{\frac{{\mathrm{x}}^2}{2}\mathrm{arcosx}+\int \frac{{\mathrm{x}}^2}{2}\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}\}={\mathrm{x}}^2\mathrm{arcosx}+\int \frac{{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\}\mathrm{we}\mathrm{have}$$$$\int \frac{{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}{=}_{\mathrm{x}=\mathrm{sint}}\int \frac{{\sin }^2\mathrm{t}}{\mathrm{cost}}\mathrm{cost}\mathrm{dt}=\int {\sin }^2\mathrm{t}\mathrm{dt}$$$$=\int \frac{1-\cos \left(2\mathrm{t}\right)}{2}\mathrm{dt}=\frac{\mathrm{t}}{2}-\frac{1}{4}\sin \left(2\mathrm{t}\right)+\mathrm{C}$$$$=\frac{\mathrm{t}}{2}-\frac{1}{2}\mathrm{sint}\mathrm{cost}+\mathrm{C}=\frac{\mathrm{arcsinx}}{2}-\frac{\mathrm{x}}{2}\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}\Rightarrow$$$$\mathrm{I}={\mathrm{x}}^2\mathrm{arcosx}+\frac{\mathrm{arcsinx}}{2}-\frac{\mathrm{x}}{2}\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}$$

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