Integration Questions

Question Number 174560 by mnjuly1970 last updated on 04/Aug/22

$$\\$$$$\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{dx}}{\:\sqrt{\mathrm{8}+\mathrm{3}{x}−\sqrt{\mathrm{1}+\left(\:{x}^{\:\mathrm{2}} +\mathrm{3}{x}\:+\mathrm{2}\right)\left({x}^{\:\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\right)}}} \\$$$$\\$$

Answered by MJS_new last updated on 04/Aug/22

$$\mathrm{1}+\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{12}\right)=\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}\right)^{\mathrm{2}} \\$$$${x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}>\mathrm{0}\forall{x}\in\left[\mathrm{0};\:\mathrm{1}\right] \\$$$$\Rightarrow \\$$$$\Omega=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\:\sqrt{−{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}}= \\$$$$\:\:\:\:\:\left[{t}=\mathrm{arcsin}\:\frac{{x}+\mathrm{1}}{\mathrm{2}}\:\rightarrow\:{dx}=\sqrt{−{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}{dt}\right] \\$$$$=\underset{\pi/\mathrm{6}} {\overset{\pi/\mathrm{2}} {\int}}{dt}=\left[{t}\right]_{\pi/\mathrm{6}} ^{\pi/\mathrm{2}} =\frac{\pi}{\mathrm{3}} \\$$

Commented by mnjuly1970 last updated on 04/Aug/22

$${thanks}\:{slot}... \\$$

Answered by MJS_new last updated on 04/Aug/22

$$\mathrm{what}'\mathrm{s}\:\mathrm{more}\:\mathrm{interesting}: \\$$$$\mho=\underset{−\mathrm{4}+\sqrt{\mathrm{3}}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{8}+\mathrm{3}{x}−\mid{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}\mid}}=? \\$$

Commented by Ar Brandon last updated on 04/Aug/22

$$\mho=\int_{−\mathrm{4}+\sqrt{\mathrm{3}}} ^{\alpha} \frac{{dx}}{\:\sqrt{\mathrm{8}+\mathrm{3}{x}+{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}}}+\int_{\alpha} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\mathrm{8}+\mathrm{3}{x}−\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}\right)}} \\$$$$\mho=\int_{−\mathrm{4}+\sqrt{\mathrm{3}}} ^{\alpha} \frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{13}}}+\int_{\alpha} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\mathrm{3}−\mathrm{2}{x}−{x}^{\mathrm{2}} }}\:,\:\alpha=\frac{−\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\$$$$\:\:\:\:=\int_{−\mathrm{4}+\sqrt{\mathrm{3}}} ^{\alpha} \frac{{dx}}{\:\sqrt{\left({x}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{3}}}+\int_{\alpha} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\mathrm{4}−\left({x}+\mathrm{1}\right)^{\mathrm{2}} }} \\$$$$\:\:\:\:=\left[\mathrm{argch}\left(\frac{{x}+\mathrm{4}}{\:\sqrt{\mathrm{3}}}\right)\right]_{−\mathrm{4}+\sqrt{\mathrm{3}}} ^{\alpha} +\left[\mathrm{arcsin}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)\right]_{\alpha} ^{\mathrm{1}} \\$$$$\:\:\:\:=\left[\mathrm{ln}\mid\left({x}+\mathrm{4}\right)+\sqrt{{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{13}}\mid\right]_{−\mathrm{4}+\sqrt{\mathrm{3}}} ^{\alpha} +\frac{\pi}{\mathrm{2}}−\mathrm{arcsin}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{4}}{\mathrm{4}}\right) \\$$