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 59576 by maxmathsup by imad last updated on 12/May/19

$$letf\left(x\right)=\int \frac{dt}{(x+t)\sqrt{t^2-x^2}}$$$$1)determineaexplicitformoff\left(x\right)$$$$2)determine\int \frac{dt}{(x+2)\sqrt{t^2-4}}and\int \frac{dt}{(x+1)\sqrt{t^2-1}}$$

Commented by tanmay last updated on 12/May/19

$$\int \frac{dt}{(t+2)\sqrt{t^2-4}}and\int \frac{dt}{(t+1)\sqrt{t^2-1}}$$$$\int \frac{dt}{(t+a)\sqrt{t^2-a^2}}$$$$k=\frac{1}{t+a}$$$$t+a=\frac{1}{k}\to dt=\frac{-dk}{k^2}$$$$\int \frac{k}{\sqrt{\left(\frac{1}{k}\right)(\frac{1}{k}-2a)}}\times \frac{-dk}{k^2}$$$$\int \frac{-dk}{k\sqrt{\frac{1-2ak}{k^2}}}$$$$p^2=1-2ak\to 2pdp=-2adk$$$$\int \frac{pdp}{a\times p}$$$$\frac{1}{a}\times p+c$$$$=\frac{\sqrt{1-2ak}}{a}+c$$$$=\frac{\sqrt{1-\frac{2a}{t+a}}}{a}+c$$$$nowputa=2$$$$\frac{\sqrt{1-\frac{4}{t+2}}}{2}+c....\left(ansforQno1\right)$$$$nowputa=1$$$$\frac{\sqrt{1-\frac{2}{t+1}}}{1}+c....\left(ansforQno2\right)$$$$$$$$$$

Commented by maxmathsup by imad last updated on 12/May/19

$$2)theQ.isdetermine\int \frac{dt}{(t+2)\sqrt{t^2-4}}and\int \frac{dt}{(t+1)\sqrt{t^2-1}}$$

Commented by maxmathsup by imad last updated on 14/May/19

$$changementt=xch\left(u\right)givef\left(x\right)=\int \frac{xsh\left(u\right)}{(x+xch\left(u\right))\mid x\mid sh\left(u\right)}du$$$$=\xi \left(x\right)\int \frac{du}{x(1+ch\left(u\right))}=\frac{\xi \left(x\right)}{x}\int \frac{du}{1+\frac{e^u+e^{-u}}{2}}=\frac{2\xi \left(x\right)}{x}\int \frac{du}{2+e^u+e^{-u}}$$$${=}_{e^u=\alpha }\frac{2\xi \left(x\right)}{x}\int \frac{1}{2+\alpha +{\alpha }^{-1}}\frac{d\alpha }{\alpha }=\frac{2\xi \left(x\right)}{x}\int \frac{d\alpha }{2\alpha +{\alpha }^2+1}$$$$=\frac{2\xi \left(x\right)}{x}\int \frac{d\alpha }{{(\alpha +1)}^2}=-\frac{2\xi \left(x\right)}{x}\frac{1}{\alpha +1}+c=-\frac{2}{x}\xi \left(x\right)\frac{1}{e^u+1}but$$$$u=argch\left(\frac{t}{x}\right)=ln(\frac{t}{x}+\sqrt{\frac{t^2}{x^2}-1})\Rightarrow e^u=\frac{t}{x}+\sqrt{\frac{t^2}{x^2}-1}\Rightarrow$$$$f\left(x\right)=-\frac{2}{x}\xi \left(x\right)\frac{1}{\frac{t}{x}+\frac{\sqrt{t^2-x^2}}{\mid x\mid }}+c=-\frac{2\xi \left(x\right)}{t+\xi \left(x\right)\sqrt{t^2-x^2}}+cwith$$$$\xi \left(x\right)=1ifx>0and\xi \left(x\right)=-1ifx<0so$$$$f\left(x\right)=\frac{-2}{t+\sqrt{t^2-x^2}}+cifx>0$$$$f\left(x\right)=\frac{2}{t-\sqrt{t^2-x^2}}+cifx<0$$

Commented by maxmathsup by imad last updated on 14/May/19

$$2)\int \frac{dt}{(t+2)\sqrt{t^2-4}}=f\left(2\right)=\frac{-2}{t+\sqrt{t^2-4}}+c$$$$\int \frac{dt}{(t+1)\sqrt{t^2-1}}=f\left(1\right)=\frac{-2}{t+\sqrt{t^2-1}}+c$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com