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cos-4-x-sinx-dx-




Question Number 11147 by suci last updated on 14/Mar/17
∫cos^4 x sinx dx=....???
cos4xsinxdx=.???
Answered by ajfour last updated on 14/Mar/17
= −((cos^5 x)/5)+C
=cos5x5+C
Commented by ajfour last updated on 14/Mar/17
let cos x= t  sin x dx = −dt  ∫cos^4 x sin x dx = −∫t^4 dt  = −(t^5 /5) +C
letcosx=tsinxdx=dtcos4xsinxdx=t4dt=t55+C
Answered by Mechas88 last updated on 17/Mar/17
∫cos4xsinxdx=  ∫(1/2)(sin(x+4x)+sin(x−4x))dx=  (1/2)∫sin5xdx+1/2∫sin−3xdx=    −((cos5x)/(10)) +(1/2)× ((cos3x)/3) +C    ((cos3x)/6) −((cos5x)/(10)) +C
cos4xsinxdx=12(sin(x+4x)+sin(x4x))dx=12sin5xdx+1/2sin3xdx=cos5x10+12×cos3x3+Ccos3x6cos5x10+C

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