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Question Number 95549 by Fikret last updated on 25/May/20

∫_(−π) ^π ∣cos^3 x∣dx

ππcos3xdx

Answered by MJS last updated on 26/May/20

∫_(−π) ^π ∣cos^3  x∣dx=4∫_0 ^(π/2) cos^3  x dx=  =∫_0 ^(π/2) (cos 3x +3cos x)dx=  =[(1/3)sin 3x +3sin x]_0 ^(π/2) =(8/3)

ππcos3xdx=4π/20cos3xdx==π/20(cos3x+3cosx)dx==[13sin3x+3sinx]0π/2=83

Commented by peter frank last updated on 26/May/20

why you change limit sir

whyyouchangelimitsir

Commented by MJS last updated on 26/May/20

because ∣cos^3  x∣ is periodic... scetch it

becausecos3xisperiodic...scetchit

Commented by peter frank last updated on 26/May/20

how abou this  ∫_(−π) ^π ∣sin x∣ dx ?

howabouthisππsinxdx?

Commented by MJS last updated on 26/May/20

∫_(−π) ^π ∣sin x∣dx=4∫_0 ^(π/2) sin x dx  the idea is to get rid of the absolute by  finding an interval where the function is  greater than or equal to zero

ππsinxdx=4π/20sinxdxtheideaistogetridoftheabsolutebyfindinganintervalwherethefunctionisgreaterthanorequaltozero

Commented by peter frank last updated on 26/May/20

thank you

thankyou

Answered by 1549442205 last updated on 26/May/20

I=∫_(−Π) ^Π ∣cos^3 x∣dx=∫_(−Π) ^((−Π)/2) (−cos^3 x)dx  +∫_(−(Π/2)) ^(Π/2) cos^3 xdx+∫_(Π/2) ^Π (−cos^3 x)dx  We have F(x)=∫cos^3 xdx=∫(1−sin^2 x)dsinx  =sinx−((sin^3 x)/3)+C.Therefore,  I=F(−Π)−F(((−Π)/2))+F((Π/2))−F(((−Π)/2))  +F((Π/2))−F(Π)=2F((Π/2))−2F(((−Π)/2))  =(4/3)−2(((−2)/3))=(8/3)(Since F(Π)=F(−Π)=0  .Thus,I=(8/3)

I=ΠΠcos3xdx=ΠΠ2(cos3x)dx+Π2Π2cos3xdx+Π2Π(cos3x)dxWehaveF(x)=cos3xdx=(1sin2x)dsinx=sinxsin3x3+C.Therefore,I=F(Π)F(Π2)+F(Π2)F(Π2)+F(Π2)F(Π)=2F(Π2)2F(Π2)=432(23)=83(SinceF(Π)=F(Π)=0.Thus,I=83

Answered by Ar Brandon last updated on 26/May/20

∫_(−π) ^π ∣cos^3 (x)∣dx=4xI; where I=∫_0 ^(π/2) cos^3 (x)dx=(2/3){Walli′s method}  ⇒∫_(−π) ^π ∣cos^3 (x)∣dx=4×(2/3)=(8/3)

ππcos3(x)dx=4xI;whereI=0π2cos3(x)dx=23{Wallismethod}ππcos3(x)dx=4×23=83

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