Question Number 42802 by maxmathsup by imad last updated on 02/Sep/18
$${calculate}\:\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{5}}{\mathrm{4}}} \:\:\:\frac{{x}^{\mathrm{3}} }{\sqrt{\mathrm{2}+{x}−{x}^{\mathrm{2}} }}{dx} \\ $$
Commented by maxmathsup by imad last updated on 05/Sep/18
![let A = ∫_(1/2) ^(5/4) (x^3 /(√(2+x−x^2 )))dx⇒A = ∫_(1/2) ^(5/4) (x^3 /(√(−x^2 +x+2)))dx we have −x^2 +x+2 =−(x^2 −x−2) =−(x^2 −2(1/2)x +(1/4)−(1/4)−2) =−{(x−(1/2))^2 −(9/4)}=(9/4) −(x−(1/2))^2 ⇒A = ∫_(1/2) ^(5/4) (x^3 /(√((9/4)−(x−(1/2))^2 )))dx =_(x−(1/2) =(3/2)sint) ∫_0 ^(π/6) ((((1/2)+(3/2)sint)^3 )/((3/2) cost)) (3/2) cost dt = ∫_0 ^(π/6) (3sint +1)^3 dt =∫_0 ^(π/6) (27sin^3 t +27sin^2 t +9sint +1)dt =27 ∫_0 ^(π/6) sin^3 t dt +27 ∫_0 ^(π/6) sin^2 t dt +9 ∫_0 ^(π/6) sint dt +(π/6) but ∫_0 ^(π/6) sint dt =[−cost]_0 ^(π/6) =1−((√3)/2) ∫_0 ^(π/6) sin^2 t dt =(1/2) ∫_0 ^(π/6) (1−cos(2t))dt=(π/(12)) −(1/4)[sin(2t)]_0 ^(π/6) =(π/(12)) −(1/4) ((√3)/2) =(π/(12)) −((√3)/8) we have sin^3 t ={((e^(it) −e^(−it) )/(2i))}^3 =−(1/(8i)){ Σ_(k=0) ^3 C_3 ^k e^(ikt) (−1)^(3−k) e^(−i(3−k)t) } =(i/8){ −e^(−i3t) +3 e^(it) e^(−i2t) − 3 e^(i2t) e^(−it) +e^(i3t) } =(i/8){e^(i3t) −e^(−i3t) −3(e^(it) −e^(−it) )} =(i/8){2i sin(3t) −6i sin(t)} =−(1/4)sin(3t) +(3/4) sin(t) ⇒ ∫_0 ^(π/6) sin^3 t dt =(3/4) ∫_0 ^(π/6) sin(t)dt−(1/4) ∫_0 ^(π/6) sin(3t)dt =−(3/4)[cost]_0 ^(π/6) +(1/(12))[ cos(3t)]_0 ^(π/6) =−(3/4)(((√3)/2)−1) +(1/(12))( −1) =(3/4) −((3(√3))/8) −(1/(12)) so the value of A is known .](Q42928.png)
$${let}\:{A}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{5}}{\mathrm{4}}} \:\:\:\:\:\:\:\frac{{x}^{\mathrm{3}} }{\sqrt{\mathrm{2}+{x}−{x}^{\mathrm{2}} }}{dx}\Rightarrow{A}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{5}}{\mathrm{4}}} \:\:\:\frac{{x}^{\mathrm{3}} }{\sqrt{−{x}^{\mathrm{2}} +{x}+\mathrm{2}}}{dx} \\ $$$${we}\:{have}\:−{x}^{\mathrm{2}} \:+{x}+\mathrm{2}\:=−\left({x}^{\mathrm{2}} −{x}−\mathrm{2}\right)\:=−\left({x}^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{x}\:+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{2}\right) \\ $$$$=−\left\{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}\right\}=\frac{\mathrm{9}}{\mathrm{4}}\:−\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:\Rightarrow{A}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{5}}{\mathrm{4}}} \:\:\:\:\:\frac{{x}^{\mathrm{3}} }{\sqrt{\frac{\mathrm{9}}{\mathrm{4}}−\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }}{dx} \\ $$$$=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{3}}{\mathrm{2}}{sint}} \:\:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:\:\:\:\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}{sint}\right)^{\mathrm{3}} }{\frac{\mathrm{3}}{\mathrm{2}}\:{cost}}\:\frac{\mathrm{3}}{\mathrm{2}}\:{cost}\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\left(\mathrm{3}{sint}\:+\mathrm{1}\right)^{\mathrm{3}} \:{dt}\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:\left(\mathrm{27}{sin}^{\mathrm{3}} {t}\:\:\:+\mathrm{27}{sin}^{\mathrm{2}} {t}\:+\mathrm{9}{sint}\:+\mathrm{1}\right){dt} \\ $$$$=\mathrm{27}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}^{\mathrm{3}} {t}\:{dt}\:+\mathrm{27}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}^{\mathrm{2}} {t}\:{dt}\:\:+\mathrm{9}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sint}\:{dt}\:+\frac{\pi}{\mathrm{6}}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sint}\:{dt}\:=\left[−{cost}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:=\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}^{\mathrm{2}} {t}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\left(\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)\right){dt}=\frac{\pi}{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \\ $$$$=\frac{\pi}{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\frac{\pi}{\mathrm{12}}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\:\:{we}\:{have}\:{sin}^{\mathrm{3}} {t}\:=\left\{\frac{{e}^{{it}} −{e}^{−{it}} }{\mathrm{2}{i}}\right\}^{\mathrm{3}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}{i}}\left\{\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{3}} \:{C}_{\mathrm{3}} ^{{k}} \:\:{e}^{{ikt}} \:\:\left(−\mathrm{1}\right)^{\mathrm{3}−{k}} \:{e}^{−{i}\left(\mathrm{3}−{k}\right){t}} \right\} \\ $$$$=\frac{{i}}{\mathrm{8}}\left\{\:\:−{e}^{−{i}\mathrm{3}{t}} \:\:+\mathrm{3}\:{e}^{{it}} \:{e}^{−{i}\mathrm{2}{t}} \:−\:\mathrm{3}\:{e}^{{i}\mathrm{2}{t}} \:{e}^{−{it}} \:\:+{e}^{{i}\mathrm{3}{t}} \right\} \\ $$$$=\frac{{i}}{\mathrm{8}}\left\{{e}^{{i}\mathrm{3}{t}} \:−{e}^{−{i}\mathrm{3}{t}} \:−\mathrm{3}\left({e}^{{it}} −{e}^{−{it}} \right)\right\}\:=\frac{{i}}{\mathrm{8}}\left\{\mathrm{2}{i}\:{sin}\left(\mathrm{3}{t}\right)\:−\mathrm{6}{i}\:{sin}\left({t}\right)\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{t}\right)\:+\frac{\mathrm{3}}{\mathrm{4}}\:{sin}\left({t}\right)\:\Rightarrow\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}^{\mathrm{3}} {t}\:{dt}\:=\frac{\mathrm{3}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}\left({t}\right){dt}−\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}\left(\mathrm{3}{t}\right){dt} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{4}}\left[{cost}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:\:+\frac{\mathrm{1}}{\mathrm{12}}\left[\:{cos}\left(\mathrm{3}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:=−\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{12}}\left(\:−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{12}}\:\:\:{so}\:{the}\:{value}\:{of}\:{A}\:{is}\:{known}\:. \\ $$