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 40153 by maxmathsup by imad last updated on 16/Jul/18

$$calculate{\int }_1^2\frac{t-2}{\sqrt{t^2-1}}dt$$

Commented by maxmathsup by imad last updated on 17/Jul/18

$$I={\int }_1^2\frac{tdt}{\sqrt{t^2-1}}dt-2{\int }_1^2\frac{dt}{\sqrt{t^2-1}}but$$$${\int }_1^2\frac{t}{\sqrt{t^2-1}}dt={\left[\sqrt{t^2-1}\right]}_1^2=\sqrt{3}alsochangementt=chugive$$$$\int \frac{dt}{\sqrt{t^2-1}}=\int \frac{shudu}{shu}=u+c=argch\left(t\right)=ln(t+\sqrt{t^2-1})+c\Rightarrow$$$${\int }_1^2\frac{dt}{\sqrt{t^2-1}}={\left[ln(t+\sqrt{t^2-1})\right]}_1^2=ln(2+\sqrt{3})\Rightarrow$$$$I=\sqrt{3}-2ln(2+\sqrt{3}).$$$$$$$$$$

Answered by sma3l2996 last updated on 16/Jul/18

$${\int }_1^2\frac{t-2}{\sqrt{t^2-1}}dt={\int }_1^2\frac{t}{\sqrt{t^2-1}}dt-2{\int }_1^2\frac{dt}{\sqrt{t^2-1}}$$$$={\left[\sqrt{t^2-1}\right]}_1^2-2{\int }_1^2\frac{t+\sqrt{t^2-1}}{(t+\sqrt{t^2-1})\sqrt{t^2-1}}dt$$$$=\sqrt{5}-2{\int }_1^2\frac{t+\sqrt{t^2-1}}{\sqrt{t^2-1}}\times \frac{dt}{t+\sqrt{t^2-1}}$$$$=\sqrt{5}-2{\int }_1^2(\frac{t}{\sqrt{t^2-1}}+1)\times \frac{1}{t+\sqrt{t^2-1}}dt$$$$=\sqrt{5}-2{\int }_1^2\frac{d(t+\sqrt{t^2-1})}{t+\sqrt{t^2-1}}$$$$=\sqrt{5}-2{[ln\mid t+\sqrt{t^2-1}\mid ]}_1^2=\sqrt{5}-2ln(2+\sqrt{5})$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com