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cos-3-x-sinx-dx-




Question Number 5262 by Kasih last updated on 03/May/16
∫cos^3 x(√(sinx))dx
cos3xsinxdx
Answered by Yozzii last updated on 03/May/16
J=∫cos^3 x(√(sinx))dx=∫cosxcos^2 x(√(sinx))dx  J=∫cosx(1−sin^2 x)(√(sinx))dx.  Let u=sinx⇒du=cosxdx.  ∴J=∫(1−u^2 )u^(1/2) du  J=∫(u^(1/2) −u^(5/2) )du  J=((2u^(3/2) )/3)−((2u^(7/2) )/7)+K  J=((2sin^(3/2) x)/3)−((2sin^(7/2) x)/7)+K
J=cos3xsinxdx=cosxcos2xsinxdxJ=cosx(1sin2x)sinxdx.Letu=sinxdu=cosxdx.J=(1u2)u1/2duJ=(u1/2u5/2)duJ=2u3/232u7/27+KJ=2sin3/2x32sin7/2x7+K

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