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 48667 by maxmathsup by imad last updated on 26/Nov/18

$$find{\int }_0^1\frac{arctan\left(x\right)}{\sqrt{1+x^2}}dx.$$

Commented by Abdo msup. last updated on 02/Dec/18

$$changementx=tantgive$$$$I={\int }_0^{\frac{\pi }{4}}\frac{t}{\sqrt{1+{tan}^{2}t}}(1+{tan}^{2}t)dt={\int }_0^{\frac{\pi }{4}}t\frac{dt}{cost}$$$$={\int }_0^{\frac{\pi }{4}}\frac{t}{cost}dt{=}_{tan\left(\frac{t}{2}\right)=u}{\int }_0^{\sqrt{2}-1}\frac{2arctanu}{\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}$$$$=4{\int }_0^{\sqrt{2}-1}\frac{arctan\left(u\right)}{1-u^2}du=4{\int }_0^{\sqrt{2}-1}arctan\left(u\right)\left(\sum _{n=0}^{\mathrm{\infty }}{u}^{2n}\right)du$$$$=4\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{\sqrt{2}-1}{u}^{2n}arctanudu=4\sum _{n=0}^{\mathrm{\infty }}{A}_{n}byparts$$$$A_n={\int }_0^{\sqrt{2}-1}{u}^{2n}arctan\left(u\right)du={\left[\frac{1}{2n+1}{u}^{2n+1}arctanu\right]}_0^{\sqrt{2}-1}$$$$+{\int }_0^{\sqrt{2}-1}\frac{1}{2n+1}{u}^{2n+1}\frac{du}{1+u^2}$$$$=\frac{1}{2n+1}{(\sqrt{2}-1)}^{2n+1}arctan(\sqrt{2}-1)+\frac{1}{2n+1}{\int }_0^{\sqrt{2}-1}\frac{u^{2n+1}}{1+u^2}du$$$$=\frac{\pi {(\sqrt{2}-1)}^{2n+1}}{8(2n+1)}+\frac{1}{2n+1}{\int }_0^{\sqrt{2}-1}\frac{u^{2n+1}}{1+u^2}dubut$$$${\int }_0^{\sqrt{2}-1}\frac{u^{2n+1}}{1+u^2}du{=}_{tan\theta =u}{\int }_0^{\frac{\pi }{8}}\frac{{tan}^{2n+1}\theta }{1+{tan}^2\theta }(1+{tan}^2\theta )d\theta$$$$={\int }_0^{\frac{\pi }{8}}{tan}^{2n+1}\theta d\theta ....becontinued....$$

Answered by Abdulhafeez Abu qatada last updated on 26/Nov/18

Terms of Service

Privacy Policy

Contact: info@tinkutara.com