How do you evaluate #int sinx/(1+cos^2x)# from #[pi/2, pi]#?

1 Answer
mason m
Jan 9, 2017

#int_(pi//2)^pisinx/(1+cos^2x)dx=pi/4#

Explanation:

#I=int_(pi//2)^pisinx/(1+cos^2x)dx#

We will make the substitution #cosx=tantheta#. Differentiating this shows that #-sinxcolor(white).dx=sec^2thetacolor(white).d theta#.

When making this substitution from #x# to #theta#, we also need to change the bounds.

  • #x=pi/2=>cos(x)=cos(pi/2)=0=tan(theta)=>theta=0#
  • #x=pi=>cos(x)=cos(pi)=-1=tan(theta)=>theta=-pi/4#

So:

#I=-int_(pi//2)^pi(-sinxcolor(white).dx)/(1+cos^2x)=-int_0^(-pi//4)(sec^2thetacolor(white).d theta)/(1+tan^2theta)#

Reversing the order of the bounds with the negative sign and using the identity #tan^2theta+1=sec^2theta#:

#I=int_(-pi//4)^0d theta=[theta]_(-pi//4)^0=0-(-pi/4)=pi/4#

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