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 10790 by Nur450737 last updated on 25/Feb/17

$${\int }_0^{\frac{\pi }{2}}\sqrt{\sin x}dx$$

Answered by bahmanfeshki last updated on 26/Feb/17

$$\sqrt{\sin x}=t$$$$t^2=\sin x\Rightarrow \cos x=\sqrt{1-t^2}$$$$2tdt=\cos xdx\Rightarrow dx=\frac{2t}{\sqrt{1-t^2}}dt$$$${\int }_0^1\frac{2t^2}{\sqrt{1-t^2}}dt={\int }_0^{\frac{\pi }{2}}2\sin {}^{2}vdv={[v-\frac{\sin 2v}{2}]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com