(i) I = \(\int\limits_0^\frac{π}{2} \frac{sinx}{1+cos^2x}dx\)
Put cosx = t ⇒ -sin xdx = dt
When x = 0 ⇒ t = cos0 = 1,

(ii) I = \(\int\limits_0^1{xe^{x^2}}dx\)
Put x2 = t ⇒ 2xdx = dt
When x = 0 ⇒ t = 0,
x = 1 ⇒ t = 1
I = \(\frac{1}{2}\)\(\int\limits_0^1e^t\)dt =
\(\Big[e^t\Big]^t_0\)
= [e1 – e0] = e – 1.

Put sin x = t ⇒ cos xdx = dt
When x = 0 ⇒ t = sin0 = 0,
