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sin-4-xdx-




Question Number 134694 by Engr_Jidda last updated on 06/Mar/21
∫sin^4 xdx
$$\int{sin}^{\mathrm{4}} {xdx} \\ $$
Answered by john_santu last updated on 06/Mar/21
J = ∫ ((1/2)+(1/2)cos 2x)^2 dx  = (1/4)∫(1+cos 2x)^2  dx  =(1/4)∫ (1+2cos 2x+(1/2)+(1/2)cos 4x)dx  =(1/4)(((3x)/2) + sin 2x + ((sin 4x)/8)) + c  = ((3x)/8) + ((sin 2x)/4) + ((sin 4x)/(32)) + c
$$\mathcal{J}\:=\:\int\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\right)^{\mathrm{2}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\right)^{\mathrm{2}} \:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{4}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\:+\:\mathrm{sin}\:\mathrm{2}{x}\:+\:\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{8}}\right)\:+\:{c} \\ $$$$=\:\frac{\mathrm{3}{x}}{\mathrm{8}}\:+\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{4}}\:+\:\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{32}}\:+\:{c}\: \\ $$

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