Question Number 66338 by mathmax by abdo last updated on 12/Aug/19
$${find}\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{5}}{\mathrm{4}}} \:\:\:\frac{{x}^{\mathrm{3}} {dx}}{\:\sqrt{\mathrm{2}+{x}−{x}^{\mathrm{2}} }} \\ $$
Commented by mathmax by abdo last updated on 14/Aug/19
![let I =∫_(1/2) ^(5/4) (x^3 /( (√(−x^2 +x+2))))dx we have −x^2 +x+2 =−(x^2 −x−2) =−((x−(1/2))^2 −2−(1/4))=−((x−(1/2))^2 −(9/4))=(9/4)−(x−(1/2))^2 changement x−(1/2)=(3/2)t give t=((2x−1)/3) ⇒ I =∫_0 ^(1/2) ((((3/2)t+(1/2))^3 )/((3/2)(√(1−t^2 ))))(3/2)dt =(1/8)∫_0 ^(1/2) (((3t+1)^3 )/( (√(1−t^2 ))))dt =(1/8) ∫_0 ^(1/2) (((3t)^3 +3(3t)^2 +3(3t)+1)/( (√(1−t^2 ))))dt=(1/8)∫_0 ^(1/2) ((27t^3 +27t^2 +9t +1)/( (√(1−t^2 ))))dt =((27)/8)∫_0 ^(1/2) (t^3 /( (√(1−t^2 ))))dt +((27)/8)∫_0 ^(1/2) (t^2 /( (√(1−t^2 ))))dt +(9/8)∫_0 ^(1/2) (t/( (√(1−t^2 ))))dt +(1/8)∫_0 ^(1/2) (dt/( (√(1−t^2 )))) we have ∫_0 ^(1/2) (t^3 /( (√(1−t^2 ))))dt =_(t=sinα) ∫_0 ^(π/6) ((sin^3 α)/(cosα))cosαdα =∫_0 ^(π/6) sinα(1−cos^2 α)dα =∫_0 ^(π/6) sinαdα +∫_0 ^(π/6) (−sinα)cos^2 αdα =[−cosα]_0 ^(π/6) +[(1/3)cos^3 α]_0 ^(π/6) =1−((√3)/2) +(1/3)((((√3)/2))^3 −1) ∫_0 ^(1/2) (t^2 /( (√(1−t^2 ))))dt =_(t=sinκ) ∫_0 ^(π/6) ((sin^2 α)/(cosα)) cosα dα =∫_0 ^(π/6) ((1−cos(2α))/2)dα =(π/(12)) −(1/4)[sin(2α)]_0 ^(π/6) =(π/(12)) −(1/4)(((√3)/2)) =(π/(12))−((√3)/8) ∫_0 ^(1/2) (t/( (√(1−t^2 ))))dt =_(t=sinα) ∫_0 ^(π/6) ((sinα)/(cosα)) cosα dα =[−cosα]_0 ^(π/6) =1−((√3)/2) ∫_0 ^(1/2) (dt/( (√(1−t^2 )))) =[arcsint]_0 ^(1/2) =(π/6) ⇒ I =((27)/8){1−((√3)/2) +(1/3)(((3(√3))/8)−1)}+((27)/8){(π/(12))−((√3)/8)}+(9/8)(1−((√3)/2)) +(π/(48))](https://www.tinkutara.com/question/Q66426.png)
$${let}\:{I}\:=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{5}}{\mathrm{4}}} \:\frac{{x}^{\mathrm{3}} }{\:\sqrt{−{x}^{\mathrm{2}} \:+{x}+\mathrm{2}}}{dx}\:\:{we}\:{have}\:−{x}^{\mathrm{2}} \:+{x}+\mathrm{2}\:=−\left({x}^{\mathrm{2}} −{x}−\mathrm{2}\right) \\ $$$$=−\left(\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}\right)=−\left(\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}\right)=\frac{\mathrm{9}}{\mathrm{4}}−\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${changement}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}{t}\:{give}\:\:{t}=\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{3}}\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}{t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }{\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\frac{\mathrm{3}}{\mathrm{2}}{dt}\:=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{\left(\mathrm{3}{t}+\mathrm{1}\right)^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\left(\mathrm{3}{t}\right)^{\mathrm{3}} \:+\mathrm{3}\left(\mathrm{3}{t}\right)^{\mathrm{2}} \:+\mathrm{3}\left(\mathrm{3}{t}\right)+\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{27}{t}^{\mathrm{3}} \:+\mathrm{27}{t}^{\mathrm{2}} \:+\mathrm{9}{t}\:+\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$=\frac{\mathrm{27}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{t}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:+\frac{\mathrm{27}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:+\frac{\mathrm{9}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$${we}\:{have}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{t}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:=_{{t}={sin}\alpha} \:\:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:\frac{{sin}^{\mathrm{3}} \alpha}{{cos}\alpha}{cos}\alpha{d}\alpha \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}\alpha\left(\mathrm{1}−{cos}^{\mathrm{2}} \alpha\right){d}\alpha\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}\alpha{d}\alpha\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \left(−{sin}\alpha\right){cos}^{\mathrm{2}} \alpha{d}\alpha \\ $$$$=\left[−{cos}\alpha\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:+\left[\frac{\mathrm{1}}{\mathrm{3}}{cos}^{\mathrm{3}} \alpha\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:=\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\left(\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:=_{{t}={sin}\kappa} \:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{sin}^{\mathrm{2}} \alpha}{{cos}\alpha}\:{cos}\alpha\:{d}\alpha\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \frac{\mathrm{1}−{cos}\left(\mathrm{2}\alpha\right)}{\mathrm{2}}{d}\alpha \\ $$$$=\frac{\pi}{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}\alpha\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:=\frac{\pi}{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:=\frac{\pi}{\mathrm{12}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:=_{{t}={sin}\alpha} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{sin}\alpha}{{cos}\alpha}\:{cos}\alpha\:{d}\alpha\:=\left[−{cos}\alpha\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \\ $$$$=\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:=\left[{arcsint}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\pi}{\mathrm{6}}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{27}}{\mathrm{8}}\left\{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}−\mathrm{1}\right)\right\}+\frac{\mathrm{27}}{\mathrm{8}}\left\{\frac{\pi}{\mathrm{12}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\right\}+\frac{\mathrm{9}}{\mathrm{8}}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:+\frac{\pi}{\mathrm{48}} \\ $$$$ \\ $$