Question

How do you find the integral of `intxcos(5x)dx`?

Answer 1

Use integration by parts.
`intudv=uv-intvdu`

Explanation:

Let `u = x`, then `du = dx, dv=cos(5x)dx, and v=1/5sin(5x)`

Substituting these values into the integration by parts equation:

`intxcos(5x)dx=x/5sin(5x)-1/5intsin(5x)dx`

The remaining integral on the right is trivial:

`intxcos(5x)dx=x/5sin(5x)+1/25cos(5x)+C`

Answer 2

`1/25(5x sin 5x + cos 5x) + C`

Explanation:

Use integration by parts `int u dv = u*v - int v du`

Let `u = x, du = dx`
Let `dv = cos 5x dx, v = int cos 5x dx`

Let `w = 5x, dw = 5 dx`
`int cos 5x dx = 1/5 int cos w dw = 1/5 sin 5x`

Fill in the integration by parts:
`int x cos 5x dx = 1/5x sin 5x - 1/5 int sin 5xdx`

To complete the last integration piece:
Let `w = 5x, dw = 5 dx`

`-1/5int sin 5xdx = -1/5 *1/5 int 5sin 5x dx = -1/25 int sin w dw = -1/25 (-cos 5x) = 1/25 cos 5x`

The final solution simplified:
`int x cos 5x dx = 1/5x sin 5x +1/25 cos 5x + C= 1/25(5x sin 5x + cos 5x) + C`