Question Number 146106 by smallEinstein last updated on 10/Jul/21
Answered by Olaf_Thorendsen last updated on 11/Jul/21
![Ω = ∫_0 ^(1/2) ((1+(√3))/( (((x+1)^2 (1−x)^6 ))^(1/4) )) dx Let x = cos2u Ω = ∫_(π/4) ^(π/6) ((1+(√3))/( (((cos2u+1)^2 (1−cos2u)^6 ))^(1/4) )) (−2sin2udu) Ω = 2∫_(π/6) ^(π/4) ((1+(√3))/( (((2cos^2 u)^2 (2sin^2 u)^6 ))^(1/4) )) sin2udu Ω = ∫_(π/6) ^(π/4) ((1+(√3))/( cosusin^3 u)) sinucosudu Ω = ∫_(π/6) ^(π/4) ((1+(√3))/( sin^2 u)) du Ω = (1+(√3))[−cotu]_(π/6) ^(π/4) Ω = (1+(√3))((√3)−1) Ω = 2](https://www.tinkutara.com/question/Q146109.png)
$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt[{\mathrm{4}}]{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{6}} }}\:{dx} \\ $$$$\mathrm{Let}\:{x}\:=\:\mathrm{cos2}{u} \\ $$$$\Omega\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{6}}} \frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt[{\mathrm{4}}]{\left(\mathrm{cos2}{u}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos2}{u}\right)^{\mathrm{6}} }}\:\left(−\mathrm{2sin2}{udu}\right) \\ $$$$\Omega\:=\:\mathrm{2}\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt[{\mathrm{4}}]{\left(\mathrm{2cos}^{\mathrm{2}} {u}\right)^{\mathrm{2}} \left(\mathrm{2sin}^{\mathrm{2}} {u}\right)^{\mathrm{6}} }}\:\mathrm{sin2}{udu} \\ $$$$\Omega\:=\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\mathrm{cos}{u}\mathrm{sin}^{\mathrm{3}} {u}}\:\mathrm{sin}{u}\mathrm{cos}{udu} \\ $$$$\Omega\:=\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\mathrm{sin}^{\mathrm{2}} {u}}\:{du} \\ $$$$\Omega\:=\:\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left[−\mathrm{cot}{u}\right]_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\Omega\:=\:\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$$$\Omega\:=\:\mathrm{2} \\ $$
Answered by gsk2684 last updated on 11/Jul/21
![∫_0 ^(1/2) ((1+(√3))/((x+1)^(1/2) (1−x)^(3/2) ))dx ∫_0 ^(1/2) ((1+(√3))/((x+1)^2 (((1−x)/(1+x)))^(3/2) ))dx, put ((1−x)/(1+x))=t⇒((−2)/((1+x)^2 ))dx=dt ∫_1 ^(1/3) ((1+(√3))/t^(3/2) )(dt/(−2))=((−1)/2)(1+(√3))[(t^((−1)/2) /((−1)/2))]_1 ^(1/3) =(1+(√3))((1/( (√(1/3))))−1)=2](https://www.tinkutara.com/question/Q146112.png)
$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx},\:{put}\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}={t}\Rightarrow\frac{−\mathrm{2}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}={dt} \\ $$$$\underset{\mathrm{1}} {\overset{\frac{\mathrm{1}}{\mathrm{3}}} {\int}}\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }\frac{{dt}}{−\mathrm{2}}=\frac{−\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left[\frac{{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\frac{−\mathrm{1}}{\mathrm{2}}}\right]_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}}−\mathrm{1}\right)=\mathrm{2} \\ $$