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Question Number 210372 by Spillover last updated on 08/Aug/24

Commented by Frix last updated on 08/Aug/24

I get π^2

Igetπ2

Answered by Berbere last updated on 08/Aug/24

∫_(−π) ^π ((2x)/(1+cos^2 (x)))dx=0  ∫_(−π) ^π ((2xsin(x))/(1+cos^2 (x)))=[−2xtan^(−1) (cos(x))]_(−π) ^π +2∫_(−π) ^π tan^(−1) (cos(x))dx_(=0)   tan^(−1) (cos(−x))dx=tan^(−1) (cos(x)  ∗  =π^2

ππ2x1+cos2(x)dx=0ππ2xsin(x)1+cos2(x)=[2xtan1(cos(x))]ππ+2ππtan1(cos(x))dx=0tan1(cos(x))dx=tan1(cos(x)=π2

Answered by klipto last updated on 09/Aug/24

  using even and odd function  ∫_(−𝛑) ^𝛑 ((2x+2xsin(x))/(1+cos^2 x))  ∫_(−𝛑) ^𝛑 ((2x)/(1+cos^2 x))dx+∫_(−𝛑) ^𝛑 ((2xsin(x))/(1+cos^2 x))dx  ∫_(−𝛑) ^𝛑 ((2x)/(1+cos^2 x))=0[because f(−x)=−f(x)]  ∫_(−𝛑) ^𝛑 ((2xsin(x))/(1+cos^2 x))dx=2∫_(−𝛑) ^𝛑 x•((sin(x))/(1+cos^2 x))  2∫udv=uv−∫_(−𝛑) ^𝛑 vdu  ∫_(−𝛑) ^𝛑 2x•((sin(x))/(1+cos^2 x))=−2xarctan(cosx)]_(−𝛑) ^𝛑 +2∫_(−𝛑) ^𝛑 arctan(cos(x))  ∫_(−𝛑) ^𝛑 2x•((sin(x))/(1+cos^2 x))=2[−xarctan(cos(x))]_(−𝛑) ^𝛑 +2∫_(−𝛑) ^𝛑 arctan(cos(x))  2∫_(−𝛑) ^𝛑 2x•((sin(x))/(1+cos^2 x))=2[−𝛑tan^(−1) (cos(𝛑)−(−𝛑arctan(cos(−𝛑))]+0  =2[(𝛑^2 /4)+(𝛑^2 /4)]=((2𝛑^2 )/2)=𝛑^2   klipto−quanta.

usingevenandoddfunctionππ2x+2xsin(x)1+cos2xππ2x1+cos2xdx+ππ2xsin(x)1+cos2xdxππ2x1+cos2x=0[becausef(x)=f(x)]ππ2xsin(x)1+cos2xdx=2ππxsin(x)1+cos2x2udv=uvππvduππ2xsin(x)1+cos2x=2xarctan(cosx)]ππ+2ππarctan(cos(x))ππ2xsin(x)1+cos2x=2[xarctan(cos(x))]ππ+2ππarctan(cos(x))2ππ2xsin(x)1+cos2x=2[πtan1(cos(π)(πarctan(cos(π))]+0=2[π24+π24]=2π22=π2kliptoquanta.

Answered by Spillover last updated on 09/Aug/24

Answered by Spillover last updated on 09/Aug/24

Answered by Spillover last updated on 09/Aug/24

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