Question

How do you find the definite integral of `5xcos(x^2)dx` from `[0, sqrtpi]`?

Answer 1

`int_0^sqrtpi5xcos(x^2)dx=0`

Explanation:

We have the integral:

`I=int_0^sqrtpi5xcos(x^2)dx`

To deal with the inside of the cosine function, let `u=x^2`. Differentiating this shows that `du=2xcolor(white).dx`. This is perfect because we have an extra `x` floating around the integrand.

`I=5/2int_0^sqrtpi2xcos(x^2)dx=5/2int_0^sqrtpiunderbrace(cos(x^2))_(cos(u))overbrace((2xcolor(white).dx))^(du)`

When converting from `x` to `u`, plug the current `x` bounds into `u=x^2`.

When `x=0`, then `u=0^2=0`. When `x=sqrtpi`, then `u=(sqrtpi)^2=pi`. Thus:

`I=5/2int_0^picos(u)du`

Since `intcos(x)dx=sin(x)`:

`I=5/2[sin(u)]_0^pi`

`I=5/2sin(pi)-5/2sin(0)`

`I=0-0`

`I=0`