Integration Questions

Question Number 90796 by  M±th+et+s last updated on 26/Apr/20

$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\$$

Commented by mathmax by abdo last updated on 26/Apr/20

$${I}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{arcsin}\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx}\:\:{cyangement}\:\:{x}={tan}\theta\:{give} \\$$$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:{arcsin}\left({sin}\left(\mathrm{2}\theta\right)\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\mathrm{2}\theta{d}\theta\:=\left[\theta^{\mathrm{2}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\:=\frac{\pi^{\mathrm{2}} }{\mathrm{9}} \\$$

Commented by  M±th+et+s last updated on 27/Apr/20

$${thanxsir}\:{but}\:{the}\:{limit}\:{has}\:{to}\:{be}\:{divide}\:{from}\: \\$$$$\mathrm{0}\:{to}\:\mathrm{1}\:{then}\:\mathrm{1}\:{to}\:\sqrt{\mathrm{3}} \\$$

Commented by mathmax by abdo last updated on 28/Apr/20

$${no}\:{sir}\:{the}\:{limits}\:{go}\:{from}\:{arctan}\left(\mathrm{0}\right)=\mathrm{0}\:{to}\:{arctan}\left(\sqrt{\mathrm{3}}\right)\:=\frac{\pi}{\mathrm{3}} \\$$

Answered by TANMAY PANACEA. last updated on 26/Apr/20

$$\int\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\$$$${x}={tana} \\$$$$\int\frac{\mathrm{1}}{{sec}^{\mathrm{2}} {a}}×{sin}^{−\mathrm{1}} \left({sin}\mathrm{2}{a}\right){sec}^{\mathrm{2}} {a}\:{da} \\$$$${a}^{\mathrm{2}} \\$$$${so}\:{answer} \\$$$$\left[\mid{tan}^{−\mathrm{1}} {x}\mid_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\right]^{\mathrm{2}} \rightarrow\left(\frac{\pi}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{9}} \\$$

Commented by  M±th+et+s last updated on 26/Apr/20

$${i}\:{think}\:{its}\:\frac{\mathrm{7}\pi^{\mathrm{2}} }{\mathrm{72}} \\$$

Answered by  M±th+et+s last updated on 28/Apr/20

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{tan}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}\right){d}\left({tan}\theta\right) \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} {cos}^{\mathrm{2}} \left(\theta\right)\:{sin}^{−\mathrm{1}} \left({sin}\mathrm{2}\theta\right)\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\theta\right)}{d}\theta \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} {sin}^{−\mathrm{1}} \left(\mathrm{2}\theta\right){d}\theta \\$$$${the}\:{last}\:{tricky}! \\$$$${sin}^{−\mathrm{1}} {sin}\theta\neq\theta,\theta>\left[\frac{\pi}{\mathrm{2}}\right]!\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\$$$${this}\:{is}\:{because}\:{sin}^{−\mathrm{1}} \left(\theta\right)\:{to}\:{be}\:{a}\:{function} \\$$$${it}\:{must}\:{defined}\:{as}\:{a}\:{mapping}\:{only} \\$$$${to}\:{sub}−{region}\:{of}\:\mathbb{R}. \\$$$$\\$$$${specifically}: \\$$$${sin}^{−\mathrm{1}} :\left[−\mathrm{1},\mathrm{1}\right]\rightarrow\left[\frac{−\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right] \\$$$$:{x}\rightarrow\theta={sin}^{−\mathrm{1}} \left({x}\right) \\$$$${therefore}.{following}\:{gives}\:{a}\:{one}\:{to}\:{one} \\$$$${mapping}: \\$$$${sin}^{−\mathrm{1}} {sin}:\left[\frac{−\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right]\rightarrow\left[\frac{−\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right] \\$$$${in}\:{our}\:{case}\:{the}\:{argument}\:{is}\:\mathrm{2}\theta\:{so}\:{our} \\$$$${range}\:{get}\:{coparessed}\:{by}\:{a}\:{factor}\:{of}\:{two} \\$$$${so}\:{we}\:{get}\:{ollowing}\:{one}\:{to}\:{many}\:{mapping}: \\$$$${sin}^{−\mathrm{1}} {sin}\left(\mathrm{2}×\_\right):\left[\frac{−\pi}{\mathrm{4}},\frac{\pi}{\mathrm{4}}\right]\rightarrow\left[\frac{−\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right] \\$$$$\\$$$${with}\:{a}\:{littile}\:{thought}\:{we}\:{can}\:{extand} \\$$$${past}\:\frac{\pi}{\mathrm{4}}\:{to}\:\frac{\pi}{\mathrm{2}}\:{and}\:{obtain}\:{following}\:{plot} \\$$$${of}\:{sin}^{−\mathrm{1}} {sin}\mathrm{2}\theta\:{with}\:\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:{which} \\$$$${is}\:{a}\:{one}\:{to}\:{many}\:{mapping}: \\$$$$\\$$$${so}\:{the}\:{integral}\:{is}\:{just}\:{the}\:{area}\:{under} \\$$$${the}\:{curve}\:{from}\:\mathrm{0}\:{to}\:\pi/\mathrm{3}\:{breaking}\:{this} \\$$$${up}: \\$$$$\Rightarrow{for}\:\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right],{the}\:{area}\:{of}\:{the}\:{triangle} \\$$$${is}\:\frac{\pi^{\mathrm{2}} }{\mathrm{16}}. \\$$$$\Rightarrow{for}\:\theta\in\left[\frac{\pi}{\mathrm{4}},\frac{\pi}{\mathrm{3}}\right],{the}\:{area}\:{trapeziod}\: \\$$$${is}\:\mathrm{5}\pi^{\mathrm{2}} /\mathrm{144} \\$$$${summing}\:{these}\:{two},{gives} \\$$$$\frac{\mathrm{7}\pi^{\mathrm{2}} }{\mathrm{72}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\$$$$\\$$