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Question Number 36406 by abdo mathsup 649 cc last updated on 01/Jun/18
find the values of I = ∫_0 ^π cos^4 dx and J = ∫_0 ^π sin^4 dx .
$${find}\:{the}\:{values}\:{of}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\pi} {cos}^{\mathrm{4}} {dx}\:{and} \\ $$$${J}\:=\:\int_{\mathrm{0}} ^{\pi} \:{sin}^{\mathrm{4}} {dx}\:. \\ $$
Commented by abdo.msup.com last updated on 05/Jun/18
we have I +J =∫_0 ^π (cos^4 x +sin^4 x)dx =∫_0 ^π {(cos^2 x +sin^2 x)^2 −2cos^2 xsin^2 x}dx =π −2 ∫_0 ^π (1/4)(sin(2x))^2 dx =π −(1/2) ∫_0 ^π ((1−cos(4x))/2)dx =π −(π/4) + (1/(16))[sin(4x)]_0 ^π =((3π)/4) also I −J = ∫_0 ^π (cos^4 x −sin^4 x)dx = ∫_0 ^π cos^2 x −sin^2 x dx = ∫_0 ^π cos(2x)dx=(1/2)[ sin(2x)]_0 ^π =0 so I =J ⇒ 2I = ((3π)/4) ⇒ I =((3π)/8) and J =((3π)/8)
$${we}\:{have}\:\:{I}\:+{J}\:=\int_{\mathrm{0}} ^{\pi} \:\left({cos}^{\mathrm{4}} {x}\:+{sin}^{\mathrm{4}} {x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \:\:\left\{\left({cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} \:−\mathrm{2}{cos}^{\mathrm{2}} {xsin}^{\mathrm{2}} {x}\right\}{dx} \\ $$$$=\pi\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}}{\mathrm{4}}\left({sin}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} {dx} \\ $$$$=\pi\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}{dx} \\ $$$$=\pi\:−\frac{\pi}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{16}}\left[{sin}\left(\mathrm{4}{x}\right)\right]_{\mathrm{0}} ^{\pi} \:=\frac{\mathrm{3}\pi}{\mathrm{4}}\:{also} \\ $$$${I}\:−{J}\:=\:\int_{\mathrm{0}} ^{\pi} \:\left({cos}^{\mathrm{4}} {x}\:−{sin}^{\mathrm{4}} {x}\right){dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \:{cos}^{\mathrm{2}} {x}\:−{sin}^{\mathrm{2}} {x}\:{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \:\:{cos}\left(\mathrm{2}{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\left[\:{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\pi} =\mathrm{0}\:{so} \\ $$$${I}\:={J}\:\Rightarrow\:\mathrm{2}{I}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:\Rightarrow\:{I}\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:{and}\:{J}\:=\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18
=(1/4)∫_0 ^Π (1+cos2x)^2 dx =(1/4)∫_0 ^Π (1+2cos2x+cos^2 2x )dx =(1/4)∫_0 ^Π dx+(1/2)∫_0 ^Π cos2xdx+(1/8)∫_0 ^Π (1+cos4x) dx =(1/4)∫_0 ^Π dx+(1/2)∫_0 ^Π cos2x dx+(1/8)∫_0 ^Π dx+(1/8)∫_0 ^Π cos4x =(Π/4)+(Π/8) =((3Π)/8)
$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\Pi} \left(\mathrm{1}+{cos}\mathrm{2}{x}\right)^{\mathrm{2}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\Pi} \left(\mathrm{1}+\mathrm{2}{cos}\mathrm{2}{x}+{cos}^{\mathrm{2}} \mathrm{2}{x}\:\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\Pi} {dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\Pi} {cos}\mathrm{2}{xdx}+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\Pi} \left(\mathrm{1}+{cos}\mathrm{4}{x}\right)\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\Pi} {dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\Pi} {cos}\mathrm{2}{x}\:{dx}+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\Pi} {dx}+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\Pi} {cos}\mathrm{4}{x} \\ $$$$=\frac{\Pi}{\mathrm{4}}+\frac{\Pi}{\mathrm{8}} \\ $$$$=\frac{\mathrm{3}\Pi}{\mathrm{8}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18
j=∫_0 ^(Π() (((1−cos2x)/2))^2 =(1/4)∫_0 ^Π 1−2cos2x+((1+cos4x)/2)dx =(1/4)∫_0 ^Π dx−(1/2)∫_0 ^Π cos2xdx+(1/8)∫_0 ^Π dx+(1/8)∫_0 ^Π cos4x =(Π/4)+(Π/8)=((3Π)/8)
$${j}=\int_{\mathrm{0}} ^{\Pi\left(\right.} \left(\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\Pi} \mathrm{1}−\mathrm{2}{cos}\mathrm{2}{x}+\frac{\mathrm{1}+{cos}\mathrm{4}{x}}{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\Pi} {dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\Pi} {cos}\mathrm{2}{xdx}+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\Pi} {dx}+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\Pi} {cos}\mathrm{4}{x} \\ $$$$=\frac{\Pi}{\mathrm{4}}+\frac{\Pi}{\mathrm{8}}=\frac{\mathrm{3}\Pi}{\mathrm{8}} \\ $$

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