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 206248 by MetaLahor1999 last updated on 10/Apr/24

$$\int \frac{1}{\sqrt{(1-t)(2-t)}}dt=...?$$

Answered by TonyCWX08 last updated on 10/Apr/24

$$\int \frac{1}{\sqrt{t^2-3t+2}}dt$$$$=\int \frac{1}{\sqrt{{(t-\frac{3}{2})}^2-\frac{1}{4}}}dt$$$$let$$$$u=t-\frac{3}{2}$$$$du=dx$$$$\int \frac{du}{\sqrt{u^2-\frac{1}{4}}}$$$$=ln(\mid u+\sqrt{u^2-\frac{1}{4}}\mid )+c$$$$=ln(\mid t-\frac{3}{2}+\sqrt{{(t-\frac{3}{2})}^2-\frac{1}{4}}\mid )+c$$$$=ln(\mid t-\frac{3}{2}+\sqrt{t^2-3t+2}\mid )+c$$$$$$

Commented by Frix last updated on 10/Apr/24

$$\mathrm{Yes}.\mathrm{But}\mathrm{for}\mathrm{the}\mathrm{less}\mathrm{advanced}\mathrm{it}\mathrm{would}\mathrm{be}$$$$\mathrm{nice}\mathrm{to}\mathrm{show}\mathrm{the}\mathrm{path}\mathrm{for}$$$$\int \frac{du}{\sqrt{u^2\pm \mathrm{c}}}=\ln \mid u+\sqrt{u^2\pm c}\mid$$

Commented by TonyCWX08 last updated on 11/Apr/24

$$Ifyouwant,sure.$$$$Butitinvolvessometrigonometrysubstitution.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com