Question

How do you evaluate the definite integral `int sin3x` from `[0,pi]`?

Answer 1

`2/3`

Explanation:

`int_0^(pi)sin3xdx`

now`d/(dx)(cos3x)=-3sin3x`

`:.int_0^(pi)sin3xdx=[-1/3cos3x]_0^pi`

`=-1/3{[cos3x]^pi-[cos3x]_0}`

`=-1/3(cos3pi-cos0)`

`=-1/3(-1-1)`

`-1/3xx-2`

`=2/3`

Answer 2

`int_0^pi sin3x = 2/3`

Explanation:

Using:

`d/dx{cosax}=-asinax`

Then:

`int_0^pi sin3x = [-1/3cos3x]_0^pi`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -1/3(cos3pi - cos 0)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -1/3(-1- 1)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2/3`