Question

Question #a888b

Answer 1

The answer is `=2`

Explanation:

We need

`intcosxdx=sinx+C`

`intsin2xdx=-1/2cos2x+C`

Therefore,

`int_0 ^(pi/2)(cosx+sin2x)=[sinx-1/2cos2x]_0^(pi/2)`

`=(sin(pi/2)-1/2cospi)-(sin0-1/2cos0)`

`=1+1/2-0+1/2`

`=2`