Question

See original problem

Answer 1

No I can't do that. I just can't figure out where the 1/25 comes from??

Answer 2

\[\frac{1}{5}x.\sin(5x) - \frac{1}{5}(-\cos(5x)) = \frac{1}{5}(x.\sin5x + \cos5x)\]

Answer 3

Original problem:
\[\int\limits_{}^{}xcos(5x) dx\]

Answer 4

u=x
du=dx
dv=cos(5x) dx
v=1/5 sin(5x)

Answer 5

You're correct setting up u, du, v, dv.
int [x cos(5x) dx] = (1/5) x sin (5x) - int [ (1/5) sin (5x)]
Since int [(1/5) sin (5x)] = - (1/25) cos (5x), the final answer is
(1/5) sin (5x) + (1/25) cos (5x) + C

Answer 6

\[\int\limits_{}^{}(1/5)\sin(5x)=(1/5)\int\limits_{}^{}(-\cos(5x))=-(1/5)(-\cos(5x))\]

Answer 7

where is the (1/25) from? I think I'm overlooking a simple alebra mistake.

Answer 8

Because the integral of cos (5x) is -(1/5) sin (5x). The reason is:
Let f(x) = cos (5x). to compute
int [ cos (5x) dx] we need to let u = 5x => du = 5 dx
int [ cos (5x) dx] = int [(1/5) cos (5x) (5 dx)' = int [ (1/5) cos (u) du]
= - (1/5) sin (u) + C = -(1/5) sin (5x) + C

Answer 9

You missed the (1/5) in front of - cos (5x). You're confused because of two -(1/5). They're different.

Answer 10

Let me work on writing it out again to catch my error. Thank you so much!