Calculus Integration by parts.

Question

S e^5xsin(x)dx

need help with steps, the book gives final answer as,

1/26(-e^(5x)Cosx+5e^(5x)sinx)+C

Answer 1

Integration by parts uses a variation of the derivative product rule:

Lets let u = sin(x)

then du = cos(x)dx

That means we need to let dv = e^5x

And the v = 1/5e^5x

So ∫sin(x)•e^5x dx = sin(x)•(1/5e^5x) - ∫1/5e^5x•cos(x)dx

Now we can repeat the process with ∫1/5e^5x•cos(x)dx first pull out the 1/5 to make:

1/5∫e^5x•cos(x)dx then let u = cos(x) making du = -sin(x)dx and dv = e^5x with v = 1/5e^5x

So 1/5∫e^5x•cos(x)dx = 1/5[cos(x)•(1/5e^5x) - ∫1/5e^5x•(-sin(x))dx] or 1/25e^5x•cos(x) + 1/25∫sin(x)•e^5xdx

Putting this together with the first portion, we get:

∫sin(x)•e^5x dx = sin(x)•(1/5e^5x) - [1/25e^5x•cos(x) + 1/25∫sin(x)•e^5xdx] or

∫sin(x)•e^5x dx = sin(x)•(1/5e^5x) - 1/25e^5x•cos(x) - 1/25∫sin(x)•e^5xdx

We can now add 1/25∫sin(x)•e^5xdx to both sides of the equation to get:

1/25∫sin(x)•e^5xdx + ∫sin(x)•e^5x dx = sin(x)•(1/5e^5x) - 1/25e^5x•cos(x) or

(1/25 + 1)∫sin(x)•e^5xdx = sin(x)•(1/5e^5x) - 1/25e^5x•cos(x) or

26/25∫sin(x)•e^5xdx = sin(x)•(1/5e^5x) - 1/25e^5x•cos(x) or

∫sin(x)•e^5xdx = 25/26[sin(x)•(1/5e^5x) - 1/25e^5x•cos(x)] or

∫sin(x)•e^5xdx = 5/26sin(x)•e^5x - 1/26e^5x•cos(x)

Factoring out 1/26 gives us: 1/26[5e^5xsin(x) - e^5xcos(x)] and then, of course, add "+C"