William W. answered • 11d

Math and science made easy - learn from a retired engineer

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^{5x}dx

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^{5x}dx] or

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

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

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

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

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

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

∫sin(x)•e^{5x}dx = 5/26sin(x)•e^{5x} - 1/26e^{5x}•cos(x)

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