Question

Evaluate : `int _0^(pi/2) "sin"^ 2  "x"  "dx"`

Answer 1

`int _0^(pi/2) "sin"^ 2  "x"  "dx"`

i = `int _0^(pi/2) (1 - cos 2"x"   "dx") /2`         [∵ 1 - 2 cos² θ = 2 sin ² θ]

`["x"/2 - ("sin"2"x")/4]_0^(pi/2)`

=`(pi/4 -("sin" pi )/4) - (0 - 0)`

=`pi/4`