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Question Number 163496 by Ahmed777hamouda last updated on 07/Jan/22

∫_0 ^(π/2) 256cos^5 ((x/2))sin^(11) ((x/2))dx

$$\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{256}\boldsymbol{{cos}}^{\mathrm{5}} \left(\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)\boldsymbol{{sin}}^{\mathrm{11}} \left(\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)\boldsymbol{{dx}} \\ $$

Answered by Ar Brandon last updated on 07/Jan/22

I=∫_0 ^(π/2) 256cos^5 ((x/2))sin^(11) ((x/2))dx=8∫_0 ^(π/2) sin^5 x(((1−cosx)/2))^3 dx =∫_0 ^(π/2) sin^5 x(1−3cosx+3cos^2 x−cos^3 x)dx =(1/2)β(3, (1/2))−(3/2)β(3, 1)+(3/2)β(3, (3/2))−(1/2)β(3, 2) =(2/5)∙(2/3)∙(2/1)−(1/2)+(3/2)∙(2/7)∙(2/5)∙(2/3)∙(2/1)−(1/(4×3×2×1)) =(8/(15))−(1/2)+(8/(35))−(1/(24))=((37)/(168))

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{256cos}^{\mathrm{5}} \left(\frac{{x}}{\mathrm{2}}\right)\mathrm{sin}^{\mathrm{11}} \left(\frac{{x}}{\mathrm{2}}\right){dx}=\mathrm{8}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{5}} {x}\left(\frac{\mathrm{1}−\mathrm{cos}{x}}{\mathrm{2}}\right)^{\mathrm{3}} {dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{5}} {x}\left(\mathrm{1}−\mathrm{3cos}{x}+\mathrm{3cos}^{\mathrm{2}} {x}−\mathrm{cos}^{\mathrm{3}} {x}\right){dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\mathrm{3},\:\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{3}}{\mathrm{2}}\beta\left(\mathrm{3},\:\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{2}}\beta\left(\mathrm{3},\:\frac{\mathrm{3}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\mathrm{3},\:\mathrm{2}\right) \\ $$$$\:\:\:=\frac{\mathrm{2}}{\mathrm{5}}\centerdot\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{2}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\centerdot\frac{\mathrm{2}}{\mathrm{7}}\centerdot\frac{\mathrm{2}}{\mathrm{5}}\centerdot\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{2}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}} \\ $$$$\:\:\:=\frac{\mathrm{8}}{\mathrm{15}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{8}}{\mathrm{35}}−\frac{\mathrm{1}}{\mathrm{24}}=\frac{\mathrm{37}}{\mathrm{168}} \\ $$

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