Working through ∫ e^x sin(x) dx for a homework set. The first round of parts leaves me with ∫ e^x cos(x) dx, and the second round hands me back the exact integral I started with. I have done it four times and I keep going in a circle. Is my choice of u wrong, or is this integral supposed to be done some other way?
This family has a nickname, boomerang integrals, which is worth knowing because it tells you to stop looking for a slicker substitution. Same trick shows up on ∫ sec³(x) dx and on ∫ e^(ax) cos(bx) dx in general. Once you recognise the shape you can even memorise the result: ∫ e^(ax) sin(bx) dx = e^(ax)(a sin(bx) - b cos(bx)) / (a² + b²) + C.
The sec³ one is nastier because the leftover piece is ∫ sec(x) dx, which you have to already know. Worth doing once by hand before you trust the formula.
Small bookkeeping thing that trips people up on this exact problem: do not carry the +C through the algebra. Solve for I first, divide by 2, then stick the constant on at the end. Otherwise you get a C/2 and start wondering whether you lost a factor somewhere.
Also just differentiate your answer. Thirty seconds of product rule tells you whether the signs are right, which matters here because two of the four terms flip.
Try u = sin(x) as a substitution instead, parts is overkill for this.
That does not close. du = cos(x) dx, and there is no cos(x) sitting in the integrand to absorb it, and the e^x cannot be rewritten in terms of sin(x). Substitution is the wrong tool here.
@abigail_ruthven · 3mo ago · 3 replies
The loop is the method, not a mistake. Name the whole thing: let
I = ∫ e^x sin(x) dx. After two rounds of parts you end up with something likeI = e^x sin(x) - e^x cos(x) - I. That is an equation, so add I to both sides and divide by 2, and you are done.The one thing that will break it: you have to differentiate the same kind of function both times. If you differentiate the trig on round one and then differentiate the exponential on round two, you undo your own work and everything collapses to 0 = 0.
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@psu_paula · 3mo ago
0 = 0 is literally what I got on attempt three and I assumed I had made a sign error. Kept both u choices as the trig function this time and it fell out in about two minutes. Thank you.
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@corwin_ashby · 3mo ago
Everybody does the 0 = 0 thing once. It is a rite of passage more than a mistake.
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