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Question Number 116998 by TANMAY PANACEA last updated on 08/Oct/20
∫_((−π)/2) ^(π/2)  ((sin^2 x)/(1+2^x ))dx
π2π2sin2x1+2xdx
Commented by Dwaipayan Shikari last updated on 08/Oct/20
∫_(−(π/2)) ^(π/2) ((sin^2 x)/(1+2^x ))dx=∫_(−(π/2)) ^(π/2) ((sin^2 x)/(1+2^(−x) ))dx  2I=∫_(−(π/2)) ^(π/2) sin^2 xdx  2I=∫_(−(π/2)) ^(π/2) ((1−cos2x)/2)  2I=(π/2)−(1/4)∫_(−(π/2)) ^(π/2) sin2x  I=(π/4)
π2π2sin2x1+2xdx=π2π2sin2x1+2xdx2I=π2π2sin2xdx2I=π2π21cos2x22I=π214π2π2sin2xI=π4
Commented by TANMAY PANACEA last updated on 08/Oct/20
thank you
thankyou
Answered by TANMAY PANACEA last updated on 08/Oct/20
∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx  ∫_(−(π/2)) ^(π/2)  ((sin^2 x)/(1+2^(−x) ))dx  2I=∫_((−π)/2) ^(π/2)  sin^2 x((1/(1+2^x ))+(2^x /(2^x +1)))dx  2I=∫_((−π)/2) ^(π/2)  ((1−cos2x)/2)dx  I=(1/4)∣x−((sin2x)/2)∣_((−π)/2) ^(π/2) =(1/4)(π)
abf(x)dx=abf(a+bx)dxπ2π2sin2x1+2xdx2I=π2π2sin2x(11+2x+2x2x+1)dx2I=π2π21cos2x2dxI=14xsin2x2π2π2=14(π)
Answered by mathmax by abdo last updated on 08/Oct/20
let∫_(−(π/2)) ^(π/2)  ((sin^2 x)/(1+2^x ))dx =I changement x=−t give  I =∫_(−(π/2)) ^(π/2)  ((sin^2 t)/(1+2^(−t) ))dt ⇒2I =∫_(−(π/2)) ^(π/2)  ((sin^2 x)/(1+2^x )) dx +∫_(−(π/2)) ^(π/2) ((sin^2 x)/(1+2^(−x) ))dx  =∫_(−(π/2)) ^(π/2) ((1/(1+2^x ))+(1/(1+2^(−x) )))sin^2 x dx =∫_(−(π/2)) ^(π/2) (((2+2^x +2^(−x) )/(1+2^(−x)  +2^x  +1)))sin^2 x dx  =∫_(−(π/2)) ^(π/2) ((1−cos(2x))/2)dx =(π/2)−(1/4)[sin(2x)]_(−(π/2)) ^(π/2)  =(π/2) ⇒2I =(π/2)  ⇒★I =(π/4) ★
letπ2π2sin2x1+2xdx=Ichangementx=tgiveI=π2π2sin2t1+2tdt2I=π2π2sin2x1+2xdx+π2π2sin2x1+2xdx=π2π2(11+2x+11+2x)sin2xdx=π2π2(2+2x+2x1+2x+2x+1)sin2xdx=π2π21cos(2x)2dx=π214[sin(2x)]π2π2=π22I=π2I=π4
Commented by TANMAY PANACEA last updated on 08/Oct/20
thank you sir
thankyousir
Commented by Bird last updated on 09/Oct/20
you are welcome
youarewelcome

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