Question Number 140447 by 676597498 last updated on 07/May/21
Answered by EDWIN88 last updated on 07/May/21
![L+M=∫_0 ^(π/2) cos^2 x sin^2 x(cos^2 x+sin^2 x)dx L+M = ∫_0 ^(π/2) cos^2 x sin^2 x dx = ∫_0 ^(π/2) cos^2 x−cos^4 x dx = ∫_0 ^(π/2) (((1+cos 2x)/2))−(((1+cos 2x)/2))^2 dx = [ ((x+(1/2)sin 2x)/2) ]_0 ^(π/2) −∫_0 ^(π/2) (((1+2cos 2x+((1+cos 4x)/2))/4))dx = ((π/4))−∫_0 ^(π/2) ((3+4cos 2x+cos 4x)/8) dx = (π/4)− [ ((3x+2sin 2x+(1/4)sin 4x)/8) ]_0 ^(π/2) = (π/4)−((3π)/(16)) = (π/(16)).](https://www.tinkutara.com/question/Q140452.png)
$$\mathrm{L}+\mathrm{M}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{L}+\mathrm{M}\:=\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}\:\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}}\right)−\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left[\:\frac{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2x}}{\mathrm{2}}\:\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} −\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\left(\frac{\mathrm{1}+\mathrm{2cos}\:\mathrm{2x}+\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{4x}}{\mathrm{2}}}{\mathrm{4}}\right)\mathrm{dx} \\ $$$$\:=\:\left(\frac{\pi}{\mathrm{4}}\right)−\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{3}+\mathrm{4cos}\:\mathrm{2x}+\mathrm{cos}\:\mathrm{4x}}{\mathrm{8}}\:\mathrm{dx} \\ $$$$\:=\:\frac{\pi}{\mathrm{4}}−\:\left[\:\frac{\mathrm{3x}+\mathrm{2sin}\:\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{4x}}{\mathrm{8}}\:\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} \\ $$$$\:=\:\frac{\pi}{\mathrm{4}}−\frac{\mathrm{3}\pi}{\mathrm{16}}\:=\:\frac{\pi}{\mathrm{16}}.\: \\ $$