Question Number 11436 by @ANTARES_VY last updated on 26/Mar/17
$$ \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\boldsymbol{\pi}}{\mathrm{4}}} {\int}}\boldsymbol{\mathrm{sinx}}Γ\boldsymbol{\mathrm{cos}}^{\mathrm{7}} \boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{dx}}. \\ $$$$\boldsymbol{\mathrm{solves}}... \\ $$
Commented by FilupS last updated on 26/Mar/17
![let u=cos(x) β΄ du=βsin(x)dx β΄β«_0 ^( Ο/4) sin(x)cos^7 (x)dx=ββ«_(x=0) ^( x=Ο/4) u^7 du ββ«_(x=0) ^( x=Ο/4) u^7 du=β[(1/8)u^8 ]_(x=0) ^(x=Ο/4) =β(1/8)[cos^8 (x)]_(x=0) ^(x=Ο/4) =β(1/8)(cos^8 ((Ο/4))βcos^8 (0)) =β(1/8)(((1/(β2)))^8 β1) =β(1/8)(2^(β(1/2)Γ8) β1) =β(1/8)(2^(β4) β1) =β(1/2^3 )((1/2^4 )β1) =β((1/2^7 )β(1/2^3 )) =β((1/2^7 )β(2^4 /2^7 )) =β(((1β2^4 )/2^7 )) =((2^4 β1)/2^7 ) =((15)/(128))](Q11442.png)
$$\mathrm{let}\:{u}=\mathrm{cos}\left({x}\right) \\ $$$$\therefore\:{du}=β\mathrm{sin}\left({x}\right){dx} \\ $$$$\: \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \mathrm{sin}\left({x}\right)\mathrm{cos}^{\mathrm{7}} \left({x}\right){dx}=β\int_{{x}=\mathrm{0}} ^{\:{x}=\pi/\mathrm{4}} {u}^{\mathrm{7}} {du} \\ $$$$β\int_{{x}=\mathrm{0}} ^{\:{x}=\pi/\mathrm{4}} {u}^{\mathrm{7}} {du}=β\left[\frac{\mathrm{1}}{\mathrm{8}}{u}^{\mathrm{8}} \right]_{{x}=\mathrm{0}} ^{{x}=\pi/\mathrm{4}} \\ $$$$=β\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{cos}^{\mathrm{8}} \left({x}\right)\right]_{{x}=\mathrm{0}} ^{{x}=\pi/\mathrm{4}} \\ $$$$=β\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{cos}^{\mathrm{8}} \left(\frac{\pi}{\mathrm{4}}\right)β\mathrm{cos}^{\mathrm{8}} \left(\mathrm{0}\right)\right) \\ $$$$=β\frac{\mathrm{1}}{\mathrm{8}}\left(\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{8}} β\mathrm{1}\right) \\ $$$$=β\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}^{β\frac{\mathrm{1}}{\mathrm{2}}Γ\mathrm{8}} β\mathrm{1}\right) \\ $$$$=β\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}^{β\mathrm{4}} β\mathrm{1}\right) \\ $$$$=β\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }β\mathrm{1}\right) \\ $$$$=β\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{7}} }β\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right) \\ $$$$=β\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{7}} }β\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{7}} }\right) \\ $$$$=β\left(\frac{\mathrm{1}β\mathrm{2}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{7}} }\right) \\ $$$$=\frac{\mathrm{2}^{\mathrm{4}} β\mathrm{1}}{\mathrm{2}^{\mathrm{7}} } \\ $$$$=\frac{\mathrm{15}}{\mathrm{128}} \\ $$