Evaluate the following integrals
Let
Let
Let x=tan t
⇒ dx=sec2tdt.
Also when x=0, t=0 and when x=1, .
Hence,
Using , we get
Let
Let 1+x2=t ⇒ 2xdx=dt.
When x=0, t=1 and when x=1, t=2.
Substituting t=1+x2
⇒ 2xdx=dt.
When t=1, x=0 and when t=2, x=1.
Hence,