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Question Number 84843 by M±th+et£s last updated on 16/Mar/20

∫((sin(7x))/(cos(3x))) dx

sin(7x)cos(3x)dx

Commented by jagoll last updated on 17/Mar/20

sin 7x = sin (4x+3x)   = sin 4x cos 3x + cos 4x sin 3x  ∫ ((sin 7x)/(cos 3x)) dx = ∫ (sin 4x + cos 4x tan 3x)dx  = −(1/4)cos 4x + ∫ cos 4x tan 3x dx  ⇒ cos 4x = cos 3x cos x − sin 3x sin x  cos 4x tan 3x = cos x −((sin^2 3x sin x)/(cos 3x))  = cos x −(((1−cos^2 3x)sin x)/(cos 3x))  = cos x−((sin x)/(cos 3x)) + cos 3x sin x  ∫ cos 4x tan 3x dx = sin x+(1/2)∫ (cos 4x+cos 2x)dx−∫ ((sin x)/(cos 3x)) dx  = sin x +(1/8)sin 4x+(1/4)sin 2x−∫ ((sin x)/(cos 3x)) dx

sin7x=sin(4x+3x)=sin4xcos3x+cos4xsin3xsin7xcos3xdx=(sin4x+cos4xtan3x)dx=14cos4x+cos4xtan3xdxcos4x=cos3xcosxsin3xsinxcos4xtan3x=cosxsin23xsinxcos3x=cosx(1cos23x)sinxcos3x=cosxsinxcos3x+cos3xsinxcos4xtan3xdx=sinx+12(cos4x+cos2x)dxsinxcos3xdx=sinx+18sin4x+14sin2xsinxcos3xdx

Commented by jagoll last updated on 17/Mar/20

∫ ((sin x cos x)/(cos 3x cos x)) dx = ∫ ((sin 2x)/(cos 4x+cos 2x)) dx  ∫ ((sin 2x)/(2cos^2 2x+cos 2x−1)) dx =  −(1/2)∫ (du/(2u^2 +u−1)) [ u = cos 2x ]  −(1/2) ∫ (du/((2u−1)(u+1)))   so easy to solve

sinxcosxcos3xcosxdx=sin2xcos4x+cos2xdxsin2x2cos22x+cos2x1dx=12du2u2+u1[u=cos2x]12du(2u1)(u+1)soeasytosolve

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