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 129158 by gowsalya last updated on 13/Jan/21

Answered by MJS_new last updated on 13/Jan/21

$$\underset{0}{\stackrel{2\pi }{\int }}\mid \cos \theta +\mathrm{i}\sin \theta -1\mid d\theta =$$$$=2\sqrt{2}\underset{0}{\stackrel{\pi }{\int }}\sqrt{1-\cos \theta }d\theta =$$$$[t=\tan \frac{\theta }{2}\to d\theta =\frac{2dt}{t^2+1}]$$$$=8\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{t}{{(t^2+1)}^{3/2}}dt=-8{\left[\frac{1}{\sqrt{t^2+1}}\right]}_0^{\mathrm{\infty }}=8$$

Commented by gowsalya last updated on 24/Jan/21

Can u elaborate step by step..ans is correct

Terms of Service

Privacy Policy

Contact: info@tinkutara.com