Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 212051 by Spillover last updated on 28/Sep/24

Answered by Spillover last updated on 28/Sep/24

Answered by Ghisom last updated on 28/Sep/24

$$\underset{0}{\stackrel{\pi /2}{\int }}\sqrt{1+\sin x}dx=$$$$[t=\tan \frac{x}{2}+\sec \frac{x}{2}]$$$$=4\underset{1}{\stackrel{1+\sqrt{2}}{\int }}\frac{t^2+2t-1}{{(t^2+1)}^2}dt=$$$$=-4{\left[\frac{t+1}{t^2+1}\right]}_1^{1+\sqrt{2}}=2$$

Commented by Spillover last updated on 28/Sep/24

$$great.thanks$$

Answered by BaliramKumar last updated on 29/Sep/24

$${\int }_0^{\frac{\pi }{2}}\sqrt{1+sinx}dx={\int }_0^{\frac{\pi }{2}}\sqrt{1+cos(\frac{\pi }{2}-x)}dx$$$${\int }_0^{\frac{\pi }{2}}\sqrt{2{cos}^2(\frac{\pi }{4}-\frac{x}{2})}dx$$$$\sqrt{2}{\int }_0^{\frac{\pi }{2}}cos(\frac{\pi }{4}-\frac{x}{2})dx$$$$-2\sqrt{2}{\left[sin(\frac{\pi }{4}-\frac{x}{2})\right]}_0^{\frac{\pi }{2}}=2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com