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Question Number 84843 by M±th+et£s last updated on 16/Mar/20
∫sin(7x)cos(3x)dx
Commented by jagoll last updated on 17/Mar/20
sin7x=sin(4x+3x)=sin4xcos3x+cos4xsin3x∫sin7xcos3xdx=∫(sin4x+cos4xtan3x)dx=−14cos4x+∫cos4xtan3xdx⇒cos4x=cos3xcosx−sin3xsinxcos4xtan3x=cosx−sin23xsinxcos3x=cosx−(1−cos23x)sinxcos3x=cosx−sinxcos3x+cos3xsinx∫cos4xtan3xdx=sinx+12∫(cos4x+cos2x)dx−∫sinxcos3xdx=sinx+18sin4x+14sin2x−∫sinxcos3xdx
∫sinxcosxcos3xcosxdx=∫sin2xcos4x+cos2xdx∫sin2x2cos22x+cos2x−1dx=−12∫du2u2+u−1[u=cos2x]−12∫du(2u−1)(u+1)soeasytosolve
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