Question

How do you integrate `int e^xsinx` by integration by parts method?

Answer 1

`int e^x sin x dx = 1/2 e^x(sin(x) - cos(x)) + C`

Explanation:

Integration by parts can be expressed:

`int u(x)v'(x) dx = u(x)v(x) - int v(x) u'(x) dx`

`color(white)()`
Let `u(x) = e^x`, `v(x) = -cos(x)`

Then `u'(x) = e^x`, `v'(x) = sin(x)`

and we find:

`int e^x sin x dx = -e^x cos(x) + int e^x cos(x) dx + C_1`

`color(white)()`
Let `u(x) = e^x`, `v(x) = sin(x)`

Then `u'(x) = e^x`, `v'(x) = cos(x)`

and we find:

`int e^x cos(x) dx = e^x sin(x) - int e^x sin(x) dx + C_2`

`color(white)()`
Combining these two results, we find:

`int e^x sin x dx = -e^x cos(x) + e^x sin(x) - int e^x sin x dx + (C_1+C_2)`

and hence:

`int e^x sin x dx = 1/2 e^x(sin(x) - cos(x)) + C`

`color(white)()`
Footnote

Integration by parts is very useful, but can end up leading you down a rabbit hole if you do not choose the parts appropriately.

In the example above, I would instead tend to find the integral by seeing what happens when you differentiate `e^x sin(x)` and `e^x cos(x)` then combine the results:

`d/(dx) e^x sin(x) = e^x sin(x) + e^x cos(x)`

`d/(dx) e^x cos(x) = e^x cos(x) - e^x sin(x)`

So by subtracting the second from the first of these, we find:

`d/(dx) (e^x (sin(x) - cos(x))) = e^x sin(x) + color(red)(cancel(color(black)(e^x(cos(x))))) - color(red)(cancel(color(black)(e^x(cos(x))))) + e^x sin(x)`

`color(white)(d/(dx) (e^x (sin(x) - cos(x)))) = 2e^x sin(x)`

Hence:

`int e^x sin(x) dx = 1/2 e^x (sin(x) - cos(x)) + C`