Evaluate the following integrals
Let
Let
Let x=tanθ
⇒ θ=tan-1x
=sin-1 (2sinθcosθ)
=sin-1 (sin2θ)
Hence f(x)=2θ
=2tan-1x
Hence
Using integration by parts, we get
-(1)
Let
Let 1+x2=t
⇒ 2x dx=dt.
Also, when x=0, t=1
and when x=1, t=2
Hence,
–(2)
Substituting value of (2) in (1), we get