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Question Number 163072 by abdullahhhhh last updated on 03/Jan/22

∫((sinx+sin3x+sin5x+sin7x)/(cosx+cos3x+cos5x+cos7x)) dx

$$\int\frac{\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{sin}}\mathrm{3}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{sin}}\mathrm{5}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{sin}}\mathrm{7}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{cosx}}+\boldsymbol{\mathrm{cos}}\mathrm{3}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{cos}}\mathrm{5}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{cos}}\mathrm{7}\boldsymbol{\mathrm{x}}}\:\boldsymbol{\mathrm{dx}} \\ $$

Answered by tounghoungko last updated on 03/Jan/22

((sin 7x+sin x+sin 5x+sin 3x)/(cos 7x+cos x+cos 5x+cos 3x)) = ((2sin 4x cos 3x+2sin 4x cos x)/(2cos 4x cos 3x+2cos 4x cos x)) = ((sin 4x)/(cos 4x)) ∫ ((d(−(1/4)cos 4x))/(cos 4x))=−(1/4)ln ∣cos 4x∣ + c

$$\:\frac{\mathrm{sin}\:\mathrm{7}{x}+\mathrm{sin}\:{x}+\mathrm{sin}\:\mathrm{5}{x}+\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{cos}\:\mathrm{7}{x}+\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{5}{x}+\mathrm{cos}\:\mathrm{3}{x}}\: \\ $$$$\:=\:\frac{\mathrm{2sin}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{3}{x}+\mathrm{2sin}\:\mathrm{4}{x}\:\mathrm{cos}\:{x}}{\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{3}{x}+\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{cos}\:{x}} \\ $$$$=\:\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{cos}\:\mathrm{4}{x}} \\ $$$$\:\int\:\frac{{d}\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{4}{x}\right)}{\mathrm{cos}\:\mathrm{4}{x}}=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{4}{x}\mid\:+\:{c}\: \\ $$

Commented by tounghoungko last updated on 03/Jan/22

sin A+sin B=2sin (((A+B)/2))cos (((A−B)/2))

$$\:\mathrm{sin}\:{A}+\mathrm{sin}\:{B}=\mathrm{2sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right) \\ $$

Commented by Ar Brandon last updated on 03/Jan/22

((2sin 4x cos 3x+2sin 4x cos x)/(2cos 4x cos 3x+2cos 4x cos x)) =((2sin4x(cos3x+cosx))/(2cos4x(cos3x+cosx)))=((sin4x)/(cos4x))

$$\frac{\mathrm{2sin}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{3}{x}+\mathrm{2sin}\:\mathrm{4}{x}\:\mathrm{cos}\:{x}}{\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{3}{x}+\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{cos}\:{x}} \\ $$$$=\frac{\mathrm{2sin4}{x}\cancel{\left(\mathrm{cos3}{x}+\mathrm{cos}{x}\right)}}{\mathrm{2cos4}{x}\cancel{\left(\mathrm{cos3}{x}+\mathrm{cos}{x}\right)}}=\frac{\mathrm{sin4}{x}}{\mathrm{cos4}{x}} \\ $$

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