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0-4-sinx-cos-7-x-dx-solves-




Question Number 11436 by @ANTARES_VY last updated on 26/Mar/17
  ∫_0 ^(𝛑/4) sinxΓ—cos^7 x dx.  solves...
$$ \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\boldsymbol{\pi}}{\mathrm{4}}} {\int}}\boldsymbol{\mathrm{sinx}}Γ—\boldsymbol{\mathrm{cos}}^{\mathrm{7}} \boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{dx}}. \\ $$$$\boldsymbol{\mathrm{solves}}… \\ $$
Commented by FilupS last updated on 26/Mar/17
let u=cos(x)  ∴ du=βˆ’sin(x)dx     ∴∫_0 ^( Ο€/4) sin(x)cos^7 (x)dx=βˆ’βˆ«_(x=0) ^( x=Ο€/4) u^7 du  βˆ’βˆ«_(x=0) ^( x=Ο€/4) u^7 du=βˆ’[(1/8)u^8 ]_(x=0) ^(x=Ο€/4)   =βˆ’(1/8)[cos^8 (x)]_(x=0) ^(x=Ο€/4)   =βˆ’(1/8)(cos^8 ((Ο€/4))βˆ’cos^8 (0))  =βˆ’(1/8)(((1/( (√2))))^8 βˆ’1)  =βˆ’(1/8)(2^(βˆ’(1/2)Γ—8) βˆ’1)  =βˆ’(1/8)(2^(βˆ’4) βˆ’1)  =βˆ’(1/2^3 )((1/2^4 )βˆ’1)  =βˆ’((1/2^7 )βˆ’(1/2^3 ))  =βˆ’((1/2^7 )βˆ’(2^4 /2^7 ))  =βˆ’(((1βˆ’2^4 )/2^7 ))  =((2^4 βˆ’1)/2^7 )  =((15)/(128))
$$\mathrm{let}\:{u}=\mathrm{cos}\left({x}\right) \\ $$$$\therefore\:{du}=βˆ’\mathrm{sin}\left({x}\right){dx} \\ $$$$\: \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \mathrm{sin}\left({x}\right)\mathrm{cos}^{\mathrm{7}} \left({x}\right){dx}=βˆ’\int_{{x}=\mathrm{0}} ^{\:{x}=\pi/\mathrm{4}} {u}^{\mathrm{7}} {du} \\ $$$$βˆ’\int_{{x}=\mathrm{0}} ^{\:{x}=\pi/\mathrm{4}} {u}^{\mathrm{7}} {du}=βˆ’\left[\frac{\mathrm{1}}{\mathrm{8}}{u}^{\mathrm{8}} \right]_{{x}=\mathrm{0}} ^{{x}=\pi/\mathrm{4}} \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{cos}^{\mathrm{8}} \left({x}\right)\right]_{{x}=\mathrm{0}} ^{{x}=\pi/\mathrm{4}} \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{cos}^{\mathrm{8}} \left(\frac{\pi}{\mathrm{4}}\right)βˆ’\mathrm{cos}^{\mathrm{8}} \left(\mathrm{0}\right)\right) \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{8}}\left(\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{8}} βˆ’\mathrm{1}\right) \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}^{βˆ’\frac{\mathrm{1}}{\mathrm{2}}Γ—\mathrm{8}} βˆ’\mathrm{1}\right) \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}^{βˆ’\mathrm{4}} βˆ’\mathrm{1}\right) \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }βˆ’\mathrm{1}\right) \\ $$$$=βˆ’\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{7}} }βˆ’\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right) \\ $$$$=βˆ’\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{7}} }βˆ’\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{7}} }\right) \\ $$$$=βˆ’\left(\frac{\mathrm{1}βˆ’\mathrm{2}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{7}} }\right) \\ $$$$=\frac{\mathrm{2}^{\mathrm{4}} βˆ’\mathrm{1}}{\mathrm{2}^{\mathrm{7}} } \\ $$$$=\frac{\mathrm{15}}{\mathrm{128}} \\ $$

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