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 26055 by abdo imad last updated on 18/Dec/17

$$calculate{\int }_0^1{(1+t^2)}^{1/2}dt$$

Answered by Joel578 last updated on 19/Dec/17

$$I={\int }_0^1\sqrt{t^2+1}dt$$$$\mathrm{Let}t=\tan x\to dt={\sec }^{2}xdx$$$$t=0\to x=0$$$$t=1\to x=\frac{\pi }{4}$$$$I={\int }_0^{\pi /4}\sqrt{{\tan }^{2}x+1}.{\sec }^{2}xdx$$$$={\int }_0^{\pi /4}{\sec }^{3}xdx$$$$\mathrm{Using}IBP\mathrm{give}:$$$$={[\frac{1}{2}\left(\sec x\tan x\right)-\frac{1}{2}\ln \mid \sec x+\tan x\mid ]}_0^{\pi /4}$$$$=[\frac{\sqrt{2}}{2}-\frac{1}{2}\ln (\sqrt{2}+1)]-[-\frac{1}{2}\ln \left(1\right)]$$$$=\frac{\sqrt{2}}{2}-\frac{1}{2}\ln (\sqrt{2}+1)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com