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Question Number 84861 by Jakir Sarif Mondal last updated on 16/Mar/20

If f(x) is an even function, then  ∫_( 0) ^π  f (cos x) dx = 2∫_( 0) ^(π/2)  f (cos x) dx

Iff(x)isanevenfunction,thenπ0f(cosx)dx=2π/20f(cosx)dx

Commented by mr W last updated on 17/Mar/20

f(x) is even, ⇒f(−x)=f(x)    ∫_0 ^π f(cos x)dx  =∫_0 ^(π/2) f(cos x)dx+∫_(π/2) ^π f(cos x)dx  =I_1 +I_2     let x=t+(π/2)  ⇒t=x−(π/2)  dx=dt  (π/2)≤x≤π ⇒0≤t≤(π/2)  cos x=cos (t+(π/2))=−cos t  f(cos x)=f(−cos t)=f(cos t)  I_2 =∫_(π/2) ^π f(cos x)dx  =∫_0 ^(π/2) f(cos t)dt  =I_1   ⇒∫_0 ^π f(cos x)dx=2I_1 =2∫_0 ^(π/2) f(cos x)dx

f(x)iseven,f(x)=f(x)0πf(cosx)dx=0π2f(cosx)dx+π2πf(cosx)dx=I1+I2letx=t+π2t=xπ2dx=dtπ2xπ0tπ2cosx=cos(t+π2)=costf(cosx)=f(cost)=f(cost)I2=π2πf(cosx)dx=0π2f(cost)dt=I10πf(cosx)dx=2I1=20π2f(cosx)dx

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