Question

How do you find the antiderivative of `e^(2x)(sin x)`?

Answer 1

`inte^(2x)sinxdx=1/5e^(2x)(2sinx-cosx)+C`

Explanation:

`I=inte^(2x)sinxdx`

We should try integration by parts. Typically when assigning values of `u` and `dv`, we want to choose a function for `u` that will get simpler as we differentiate it. However, we see that `e^(2x)` will stay being an exponential function and `sinx` will bounce back and forth through trigonometric functions.

In fact, we see it doesn't really matter which we choose for `u` and which for `dv`. On a whim, let:

`{(u=e^(2x),=>,du=2e^(2x)dx),(dv=sinxdx,=>,v=-cosx):}`

Then:

`I=uv-intvdu`

`I=-e^(2x)cosx-int(-cosx)(2e^(2x)dx)`

`I=-e^(2x)cosx+2inte^(2x)cosxdx`

Perform integration by parts once more. Again choose `e^(2x)` as `u`.

`{(u=e^(2x),=>,du=2e^(2x)dx),(dv=cosxdx,=>,v=sinx):}`

`I=-e^(2x)cosx+2[uv-intvdu]`

`I=-e^(2x)cosx+2uv-2intvdu`

`I=-e^(2x)cosx+2e^(2x)sinx-2intsinx(2e^(2x)dx)`

`I=-e^(2x)cosx+2e^(2x)sinx-4inte^(2x)sinxdx`

Notice that we have the integral we started out with on both sides of the equation now--that is, we can write:

`I=-e^(2x)cosx+2e^(2x)sinx-4I`

Solve for `I` treating the entire integral like we would any other variable:

`5I=-e^(2x)cosx+2e^(2x)sinx`

`5I=e^(2x)(2sinx-cosx)`

`I=1/5e^(2x)(2sinx-cosx)`

`inte^(2x)sinxdx=1/5e^(2x)(2sinx-cosx)*C`

Answer 2

`int \ e^(2x)sin(x) \ dx = e^(2x)(2/5sinx- 1/5cosx) + c`

Explanation:

Another approach when dealing with integrals of the form:

`I_1 = int \ e^(ax)sin(omega x) \ \`, or `\ \ I_2 = int \ e^(ax)cos(omega x)`

Is to use some intuition to determine the form of the solution.

Irrespective of the trig function, the results are analogous, so wlog let us consider only `I_1`. If we use Integration by parts we find that we can decompose `I_1` as follows:

`I_1 = A_1e^(ax)cos(omega x) + A_2 \ int \ e^(ax)cos(omega x)`

Which doesn't help much until we apply Integration By Parts a second time, giving:

`I_1 = A_4e^(ax)sin(omega x) + A_5e^(ax)cos(omega x) + A_6 \ int \ e^(ax)sin(omega x)`

Or:

`I_1 = A_4e^(ax)sin(omega x) + A_5e^(ax)cos(omega x) + A_6 I_1`

Which is now an algebraic equation which can be solved for `I_1`

`I_1 = e^(ax)(A_7sin(omega x) + A_8cos(omega x))`

Knowing this, we can start with an assumption of the integral result and differentiate it to see if it works, thus:

Assume a solution of the form:

`y= e^(2x)(Asinx + Bcosx)`

Differentiating wrt `x` and applying the product rule we get:

`dy/dx = e^(2x)(Acosx-Bsinx) + 2e^(2x)(Asinx + Bcosx)`
`" " = e^(2x)((A+2B)cosx+(2A-B)sinx)`

We want `dy/dx = e^(2x)sinx`, so equating coefficients of sine and cosine we get:

`cosx: \ A+2B=0`
`sinx: \ 2A-B=1`

Solving these simultaneous equations we get

`A=2/5, \ B=-1/5`

Thus we have:

`y= e^(2x)(2/5sinx- 1/5cosx)`

Hence as this is an antiderivative of our initial integral, and we have:

`int \ e^(2x)sin(x) \ dx = e^(2x)(2/5sinx- 1/5cosx) + c`