Math
posted by Jessica .
Calculate the integrals by partial fractions and using the indicated substitution. Show the results you get are the same.
dx/1x^2; substitution x= sin pheta
I understand how to do the partial fraction part, but not the second part and I don't know how they are similar. Any help would be appreciated on what to do

They want you to do it two ways. The first is to change it to
dx/[2(1+x)] + dx/[2(1x)]
which integrates to
(1/2)[ln(1+x)  ln(1x)]
= (1/2)ln[(1+x)/(1x)]
In the substitution method, with x = sin u
dx = cos u du
Integral dx/(1x^2)= cos u du/1sin^2u
= Integral du/cos u = Integral (sec u)
= (1/2)log[(1+sinu)/(1sinu)]
= (1/2)log[(1+x)/(1x)]