Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 84666 by jagoll last updated on 14/Mar/20

$$\int x{\sin }^{-1}\left(x\right)dx$$

Commented by mathmax by abdo last updated on 15/Mar/20

$$A=\int xarcsinxdxbyparts{u}^{{}^{\prime }}=xandv=arcsinx\Rightarrow$$$$A=\frac{x^2}{2}arcsinx-\int \frac{x^2}{2}\frac{dx}{\sqrt{1-x^2}}wehave$$$$\int \frac{x^{2}dx}{\sqrt{1-x^2}}=\int \frac{x^2-1+1}{\sqrt{1-x^2}}dx=\int \sqrt{1-x^2}+\int \frac{dx}{\sqrt{1-x^2}}$$$$=-\int \sqrt{1-x^2}dx+arcsinx+Cbut$$$$\int \sqrt{1-x^2}dx{=}_{x=sint}\int costcostdt=\int \frac{1+cos\left(2t\right)}{2}dt$$$$=\frac{1}{2}t+\frac{1}{4}sin\left(2t\right)=\frac{1}{2}t+\frac{1}{2}sintcost=\frac{1}{2}arcsinx+\frac{1}{2}x\sqrt{1-x^2}$$$$\Rightarrow A=\frac{1}{2}{x}^{2}arcsinx-\frac{1}{2}(arcsinx-\frac{1}{2}arcsinx-\frac{1}{2}x\sqrt{1-x^2})+C$$$$=\frac{x^2}{2}arcsinx-\frac{1}{4}arcsinx+\frac{1}{4}x\sqrt{1-x^2}+C$$$$=(\frac{x^2}{2}-\frac{1}{4})arcsinx-\frac{x}{4}\sqrt{1-x^2}+C$$$$$$

Commented by mathmax by abdo last updated on 15/Mar/20

$$A=(\frac{x^2}{2}-\frac{1}{4})arcsinx+\frac{x}{4}\sqrt{1-x^2}+C$$

Answered by MJS last updated on 15/Mar/20

$$\int x\arcsin xdx=$$$$\mathrm{by}\mathrm{parts}$$$$u=\arcsin x\to u^{\prime }=\frac{1}{\sqrt{1-x^2}}$$$$v^{\prime }=x\to v=\frac{1}{2}{x}^2$$$$=\frac{1}{2}{x}^2\arcsin x-\frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}}dx=$$$$$$$$\int \frac{x^2}{\sqrt{1-x^2}}dx=$$$$[t=\arcsin x\to dx=\cos tdt]$$$$=\int {\sin }^{2}tdt=\frac{1}{4}(2t-\sin 2t)=$$$$=\frac{1}{2}(\arcsin x-x\sqrt{1-x^2})$$$$$$$$=\frac{1}{4}((2x^2-1)\arcsin x+x\sqrt{1-x^2})+C$$

Commented by jagoll last updated on 15/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com