Question Number 73689 by ajfour last updated on 14/Nov/19
$$\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \left(\mathrm{2}+{x}\right)\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\right){dx}\:=\:? \\ $$
Answered by mind is power last updated on 15/Nov/19
$$\frac{{d}}{{dx}}\left\{{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\right)\right\}={f}\left({x}\right)^{'} =\left\{\frac{−\mathrm{3}{x}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}\:−\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\left(\mathrm{2}+{x}\right)^{\mathrm{2}} }\right\}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }{\left(\mathrm{2}+{x}\right)^{\mathrm{2}} }}} \\ $$$$=\left\{\frac{−\mathrm{3}{x}\left(\mathrm{2}+{x}\right)−\left(\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} \right)}{\left(\mathrm{2}+{x}\right)^{\mathrm{2}} .\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}\right\}.\frac{\mathrm{2}+{x}}{\:\sqrt{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }}=\left\{\frac{−\mathrm{6}{x}−\mathrm{3}}{\mid\mathrm{2}{x}+\mathrm{1}\mid}\right\}.\frac{\mathrm{1}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }} \\ $$$$=\frac{−\mathrm{3}{sign}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}\:\:\:{if}\:{x}\epsilon\left[−\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right] \\ $$$$=\frac{−\mathrm{3}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }},\frac{\mathrm{3}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}\:\:{ifx}\in\left[−\mathrm{1},\frac{−\mathrm{1}}{\mathrm{2}}\right] \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\mathrm{2}+{x}\right){sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\right)=\left[\left(\mathrm{2}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right){sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} −\int_{−\mathrm{1}} ^{−\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{4}{x}+{x}^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{3}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}−\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{4}{x}+{x}^{\mathrm{2}} }{\mathrm{2}}.\frac{−\mathrm{3}{dx}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }} \\ $$$${all}\:{we}\:{need}\:{is}\:\int\frac{\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right){dx}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }},{x}={sin}\left(\theta\right),\theta\in\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right]\Rightarrow{cos}\left(\theta\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\int\frac{\mathrm{3}\left({sin}^{\mathrm{2}} +\mathrm{4}{sin}\left(\theta\right)\right){cos}\left(\theta\right){d}\theta}{\left\{\mathrm{2}+{sin}\left(\theta\right)\right\}.{cos}\left(\theta\right).\sqrt{\mathrm{3}}}=\int\frac{\mathrm{3}\left({sin}\left(\theta\right)+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{12}}{\:\sqrt{\mathrm{3}}\left(\mathrm{2}+{sin}\left(\theta\right)\right\}}{d}\theta \\ $$$$=\sqrt{\mathrm{3}}\int\left({sin}\left(\theta\right)+\mathrm{2}\right)−\mathrm{4}\sqrt{\mathrm{3}}\int\frac{{d}\theta}{\mathrm{2}+{sin}\left(\theta\right)}\:\:\:,{tg}\left(\frac{\theta}{\mathrm{2}}\right)={t}\:{in}\mathrm{2}{nd} \\ $$$$=−\sqrt{\mathrm{3}}{cos}\left(\theta\right)+\mathrm{2}\sqrt{\mathrm{3}}\theta−\mathrm{4}\sqrt{\mathrm{3}}\int\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=−\sqrt{\mathrm{3}}{cos}\left(\theta\right)+\mathrm{2}\sqrt{\mathrm{3}\:\theta}\:−\mathrm{4}\sqrt{\mathrm{3}}\int\frac{{dt}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}} \\ $$$$=−\sqrt{\mathrm{3}}{cos}\left(\theta\right)+\mathrm{2}\sqrt{\mathrm{3}}\theta−\mathrm{8}{arctan}\left(\left(\frac{\mathrm{2}{tg}\left(\frac{\theta}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$$$\int\frac{\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right){dx}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}=−\sqrt{\mathrm{3}}\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\:+\mathrm{2}\sqrt{\mathrm{3}}{arcsin}\left({x}\right)−\mathrm{8}{arctan}\left(\frac{\mathrm{2}{tg}\left(\frac{{arcsin}\left({x}\right)}{\mathrm{2}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right). \\ $$$$\left.−\left\{−\sqrt{\mathrm{3}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}\frac{\pi}{\mathrm{6}}−\mathrm{8}.\frac{\pi}{\mathrm{12}}.\right)\right\}+\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{2}\sqrt{\mathrm{3}}.\frac{\pi}{\mathrm{2}}+\mathrm{8}.\frac{\pi}{\mathrm{6}}\right\}+\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}\sqrt{\mathrm{3}}.\frac{\pi}{\mathrm{2}}−\mathrm{8}.\frac{\pi}{\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\pi}{\:\sqrt{\mathrm{3}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by ajfour last updated on 15/Nov/19
$${Thanks},\:\mathcal{P}{owerful}\:\mathcal{M}{ind}! \\ $$
Commented by mind is power last updated on 15/Nov/19
$${y}\:'{re}\:{welcom} \\ $$
Answered by MJS last updated on 15/Nov/19
$$\int\left(\mathrm{2}+{x}\right)\mathrm{arcsin}\:\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arccos}\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+\mathrm{2}}\:\rightarrow\:{dx}=−\frac{\left({x}+\mathrm{2}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{3}}{dt}\right] \\ $$$$\left(\mathrm{1}\right) \\ $$$$\mathrm{for}\:\mathrm{0}\leqslant{t}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$=−\mathrm{9}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\begin{bmatrix}{\int{u}'{v}={uv}−\int{uv}'}\\{{u}'=\frac{\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }\:\Rightarrow\:{u}=−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }}\\{{v}={t}\:\Rightarrow\:{v}'=\mathrm{1}}\end{bmatrix} \\ $$$$=\frac{\mathrm{9}{t}}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }−\frac{\mathrm{9}}{\mathrm{2}}\int\frac{{dt}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }= \\ $$$$ \\ $$$$\:\:\:\:\:\int\frac{{dt}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{w}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:\rightarrow\:{dt}=\frac{\mathrm{2}{dw}}{{w}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$\:\:\:\:\:=\mathrm{2}\int\frac{{w}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{3}{w}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dw}= \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}{w}}{\mathrm{3}\left(\mathrm{3}{w}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\sqrt{\mathrm{3}}{w}\:= \\ $$$$\:\:\:\:\:=\frac{\mathrm{sin}\:{t}}{\mathrm{3}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\left(\sqrt{\mathrm{3}}\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$=\frac{\mathrm{9}{t}}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }−\frac{\mathrm{3sin}\:{t}}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)}−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{arctan}\:\left(\sqrt{\mathrm{3}}\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{for}\:\frac{\pi}{\mathrm{2}}\leqslant{t}\leqslant\pi \\ $$$$=−\mathrm{9}\pi\int\frac{\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }{dt}+\mathrm{9}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }{dt}= \\ $$$$=\frac{\mathrm{9}\pi}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }−\left(\mathrm{the}\:\mathrm{above}\right) \\ $$$$ \\ $$$$\Rightarrow \\ $$$$ \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{2}+{x}\right)\mathrm{arcsin}\:\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\:{dx}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\pi+\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 15/Nov/19
$$\mathcal{S}{uperb}\:{Sir};\:{thanks}\:\left({you}\:{are}\right. \\ $$$$\left.{indeed}\:{an}\:{integral}\:{lover}\right)! \\ $$
Commented by ajfour last updated on 15/Nov/19
$${Sir}\:{what}\:{if} \\ $$$${I}=\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{2}+{x}\right)^{\mathrm{2}} \mathrm{arcsin}\:\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\:{dx}\:\:? \\ $$
Commented by MJS last updated on 15/Nov/19
$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{get} \\ $$$$−\mathrm{27}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}\:\mathrm{and}\:\left(−\mathrm{27}\int\frac{\left(\pi−{t}\right)\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}\right) \\ $$$$\mathrm{same}\:\mathrm{path}\:\mathrm{leads}\:\mathrm{to} \\ $$$$−\mathrm{27}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}= \\ $$$$=\frac{\mathrm{9}{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }+\frac{\mathrm{3sin}\:{t}\:\left(\mathrm{5}−\mathrm{2cos}\:{t}\right)}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }+\mathrm{3}\sqrt{\mathrm{3}}\mathrm{arctan}\:\left(\sqrt{\mathrm{3}}\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\right) \\ $$$$−\mathrm{27}\int\frac{\left(\pi−{t}\right)\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}= \\ $$$$=−\mathrm{27}\pi\int\frac{\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}+\mathrm{27}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }= \\ $$$$=\frac{\mathrm{9}\pi}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }−\left(\mathrm{the}\:\mathrm{above}\right) \\ $$$$\Rightarrow \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{2}+{x}\right)^{\mathrm{2}} \mathrm{arcsin}\:\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\:{dx}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\pi+\frac{\mathrm{15}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 15/Nov/19
$$\mathcal{G}{reat}\:,\:{Sir}.\:{Absolutely}\:{correct}, \\ $$$${I}\:{applied}\:{it},\:{thanks}. \\ $$
Commented by MJS last updated on 15/Nov/19
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$