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cos-4-x-sinx-dx-

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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=−dt∫cos4xsinxdx=−∫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(x−4x))dx=12∫sin5xdx+1/2∫sin−3xdx=−cos5x10+12×cos3x3+Ccos3x6−cos5x10+C

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