Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 134987 by 0731619177 last updated on 09/Mar/21

Answered by Olaf last updated on 09/Mar/21

$${\mathrm{I}}_2\left(k\right)={\int }_0^{\frac{\pi }{2}}\frac{x\sin x\cos x}{\sqrt{1-k^2{\sin }^{2}x}}dx$$$${\mathrm{I}}_2\left(k\right)=-\frac{1}{k^2}{\int }_0^{\frac{\pi }{2}}x\left[\frac{-2k^2\sin x\cos x}{2\sqrt{1-k^2{\sin }^{2}x}}\right]dx$$$${\mathrm{I}}_2\left(k\right)=-\frac{1}{k^2}{\left[x\sqrt{1-k^2{\sin }^{2}x}\right]}_0^{\frac{\pi }{2}}$$$$+\frac{1}{k^2}{\int }_0^{\frac{\pi }{2}}\sqrt{1-k^2{\sin }^{2}x}dx$$$${\mathrm{I}}_2\left(k\right)=-\frac{\pi \sqrt{1-k^2}}{2k^2}+\frac{\mathrm{E}\left(k\right)}{k^2}$$$$\mathrm{Let}{k}^{\prime }=\sqrt{1-k^2}$$$${\mathrm{I}}_2\left(k\right)=-\frac{\pi k^{\prime }}{2k^2}+\frac{\mathrm{E}\left(k\right)}{k^2}$$

Commented by 0731619177 last updated on 09/Mar/21

$$tankssir$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com