Question

How do you integrate `\int _ { 0} ^ { \pi } \cos ( 3x ) d x`?

Answer 1

The answer is `=0`

Explanation:

We need

`intcosxdx=sinx+C`

Therefore,

`int_0^picos(3x)dx=[sin(3x)/3]_0^pi`

`=(sin(3pi)/3-sin0/3)`

`=0-0`