Posted by Jessica on .
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

Math 
drwls,
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)]