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 82573 by Power last updated on 22/Feb/20

Commented by Tony Lin last updated on 22/Feb/20

$$letx=asec\theta ,dx=asec\theta tan\theta d\theta$$$$\Rightarrow sec\theta =\frac{x}{a},tan\theta =\frac{\sqrt{x^2-a^2}}{a}$$$$a^2\int {tan}^2\theta sec\theta d\theta$$$$=a^2\int ({sec}^2\theta -1)sec\theta d\theta$$$$=a^2[\int {sec}^3\theta d\theta -\int sec\theta d\theta ]$$$$\int {sec}^{3}d\theta$$$$=sec\theta tan\theta -\int {tan}^2\theta sec\theta d\theta$$$$=sec\theta tan\theta -\int ({sec}^2\theta -1)sec\theta d\theta$$$$=sec\theta tan\theta -\int {sec}^{3}d\theta +\int sec\theta d\theta$$$$\Rightarrow \int {sec}^{3}d\theta$$$$=\frac{1}{2}(sec\theta tan\theta +ln\mid sec\theta +tan\theta \mid )+c$$$$\therefore a^2\int {tan}^2\theta sec\theta d\theta$$$$=\frac{1}{2}{a}^2(sec\theta tan\theta -ln\mid sec\theta +tan\theta \mid )+c$$$$=\frac{1}{2}(x\sqrt{x^2-a^2}-aln\mid x+\sqrt{x^2-a^2}\mid )+c$$

Commented by mathmax by abdo last updated on 22/Feb/20

$$A=\int \sqrt{x^2-a^2}dxchangementx=ach\left(t\right)give$$$$A=\int \mid a\mid sh\left(t\right)ash\left(t\right)dt=a\mid a\mid \int {sh}^{2}tdt$$$$=a\mid a\mid \int \frac{ch\left(2t\right)-1}{2}=\frac{a\mid a\mid }{2}\int ch\left(2t\right)dt-\frac{a\mid a\mid t}{2}$$$$=\frac{a\mid a\mid }{4}sh\left(2t\right)-\frac{a\mid a\mid }{2}t+c=\frac{a\mid a\mid }{2}sh\left(t\right)ch\left(t\right)-\frac{a\mid a\mid }{2}t+c$$$$=\frac{a\mid a\mid }{2}\sqrt{\frac{x^2}{a^2}-1}\times \frac{x}{a}-\frac{a\mid a\mid }{2}argch\left(\frac{x}{a}\right)+C$$$$=\frac{a\mid a\mid }{2\mid a\mid a}x\sqrt{x^2-a^2}-\frac{a\mid a\mid }{2}ln(\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1})+C$$$$=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a\mid a\mid }{2}ln(\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{\mid a\mid })+C$$

Commented by Power last updated on 22/Feb/20

$$\mathrm{thanks}$$

Commented by msup trace by abdo last updated on 22/Feb/20

$$youarewelcome$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com