Question

How do you evaluate the definite integral `int sin2theta` from `[0,pi/6]`?

Answer 1

`int_0^(pi/6)sin2theta=1/4`

Explanation:

`int_0^(pi/6)sin(2theta)d theta`

let
`color(red)(u=2theta)`
`color(red)(du=2d theta)`
`color(red)(d theta=(du)/2)`
The boundaries are changed to `color(blue)([0,pi/3])`

`int_0^(pi/6)sin2thetad theta`
`=int_color(blue)0^color(blue)(pi/3)sincolor(red)(u(du)/2)`
`=1/2int_0^(pi/3)sinudu`
As we know the`intsinx=-cosx`

`=-1/2(cos(pi/3)-cos0)`
`=-1/2(1/2-1)=-1/2*-1/2=1/4`

therefore,`int_0^(pi/6)sin2theta=1/4`