sin-2-x-cos-4-x-dx-

Question Number 140571 by liberty last updated on 09/May/21

$$\int\:\left(\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}\:\right)\mathrm{dx}\:=? \\ $$

Answered by bramlexs22 last updated on 09/May/21

Answered by physicstutes last updated on 09/May/21

I = ∫(sin^2 x + cos^4 x)dx I_1 = ∫sin^2 xdx = (1/2)∫(1−cos2x)dx = (1/2)(x−(1/2)sin 2x) + A I_2 = ∫cos^4 xdx = ∫[(1/2)(1+cos 2x)]^2 dx = (1/4)∫ (1+ 2 cos 2x + cos^2 2x)dx = (1/2)∫(1+2cos 2x + (1/2)(1+cos 4x))dx = (1/2)(x + sin 2x + (1/2)x +(1/8)sin 4x)+B ∴ I = (1/2)x − (1/4)sin 2x +(1/2)x+(1/2)sin 2x + (1/4)x + (1/(16))sin 4x + k I = (5/4)x−(1/4)sin 2x + (1/(16))sin 4x + k

$${I}\:=\:\int\left(\mathrm{sin}^{\mathrm{2}} {x}\:+\:\mathrm{cos}^{\mathrm{4}} {x}\right){dx} \\ $$$${I}_{\mathrm{1}} \:=\:\int\mathrm{sin}^{\mathrm{2}} {xdx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}−\mathrm{cos2}{x}\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\right)\:+\:{A} \\ $$$${I}_{\mathrm{2}} \:=\:\int\mathrm{cos}^{\mathrm{4}} {xdx}\:=\:\int\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\right)\right]^{\mathrm{2}} {dx}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\mathrm{1}+\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{x}\:+\:\mathrm{cos}^{\mathrm{2}} \mathrm{2}{x}\right){dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{2}{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{4}{x}\right)\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}\:+\:\mathrm{sin}\:\mathrm{2}{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}\:\mathrm{4}{x}\right)+{B} \\ $$$$\therefore\:{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}{x}\:+\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\:+\:\frac{\mathrm{1}}{\mathrm{4}}{x}\:+\:\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{4}{x}\:+\:{k} \\ $$$${I}\:=\:\frac{\mathrm{5}}{\mathrm{4}}{x}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}{x}\:+\:\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{4}{x}\:+\:{k} \\ $$

Answered by EDWIN88 last updated on 09/May/21

∫ (sin^2 x+cos^2 x(1−sin^2 x) dx = ∫ (sin^2 x+cos^2 x−(1/4)sin^2 2x)dx =∫ (1−(1/4)(((1−cos 4x)/2)))dx =∫ ((7/8)+((cos 4x)/8)) dx = ((7x)/8)+((cos 4x)/(32)) + c = ((28x+cos 4x)/(32)) + c

$$\int\:\left(\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)\:\mathrm{dx}\right. \\ $$$$=\:\int\:\left(\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x}\right)\mathrm{dx} \\ $$$$=\int\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{4x}}{\mathrm{2}}\right)\right)\mathrm{dx} \\ $$$$=\int\:\left(\frac{\mathrm{7}}{\mathrm{8}}+\frac{\mathrm{cos}\:\mathrm{4x}}{\mathrm{8}}\right)\:\mathrm{dx}\:=\:\frac{\mathrm{7x}}{\mathrm{8}}+\frac{\mathrm{cos}\:\mathrm{4x}}{\mathrm{32}}\:+\:\mathrm{c} \\ $$$$=\:\frac{\mathrm{28x}+\mathrm{cos}\:\mathrm{4x}}{\mathrm{32}}\:+\:\mathrm{c}\: \\ $$

Answered by mathmax by abdo last updated on 09/May/21

I=∫ (sin^2 x +cos^4 x)dx ⇒I =∫ sin^2 x dx +∫ cos^4 x dx =∫ ((1−cos(2x))/2)dx +∫ (((1+cos(2x))/2))^2 dx =(x/2)−(1/4)sin(2x) +(1/4)∫ (1+2cos(2x)+cos^2 (2x))dx =(x/2)−(1/4)sin(2x) +(x/4) +(1/4)sin(2x) +(1/4)∫ ((1+cos(4x))/2)dx =((3x)/4) +(x/8) +(1/(32))sin(4x) +C =((7x)/8) +(1/(32))sin(4x) +C

$$\mathrm{I}=\int\:\left(\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:+\mathrm{cos}^{\mathrm{4}} \mathrm{x}\right)\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:+\int\:\mathrm{cos}^{\mathrm{4}} \mathrm{x}\:\mathrm{dx} \\ $$$$=\int\:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}\mathrm{dx}\:+\int\:\left(\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \:\mathrm{dx} \\ $$$$=\frac{\mathrm{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{2x}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2x}\right)\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)\:+\frac{\mathrm{x}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{4x}\right)}{\mathrm{2}}\mathrm{dx} \\ $$$$=\frac{\mathrm{3x}}{\mathrm{4}}\:+\frac{\mathrm{x}}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{sin}\left(\mathrm{4x}\right)\:+\mathrm{C} \\ $$$$=\frac{\mathrm{7x}}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{sin}\left(\mathrm{4x}\right)\:+\mathrm{C} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply