Calculus: integration by parts: ∫exsinxdx
The tutor shows how to integrate exsinxdx by parts.
I’ve written a couple of posts on integration by parts: here and here. The method is used on products, and depends on choosing one function to integrate, then differentiating the other:
∫uv’=uv-∫vu’
With one of the functions xk for some integer k, that one will typically be differentiated, since it will eventually disappear (after k iterations). For k>2, the process will perhaps be long and messy, with lots of chances for error, but likely possible nonetheless – especially if the other function is integrable.
What if neither function will disappear, no matter how many times it’s differentiated? Today’s example is such a case:
∫exsinxdx
Happily, ex is easily integrable. Let’s imagine it as ∫(sinx)exdx: we’ll integrate exdx to ex, then differentiate sinx to cosx:
∫uv’=uv-∫vu’
becomes
∫(sinx)exdx=(sinx)ex -∫excosxdx
For the second round, we’ll integrate exdx again, while differentiating cosx:
∫excosxdx=(cosx)ex – ∫ex(-sinx)dx
⇒∫excosxdx=(cosx)ex + ∫ex(sinx)dx
Notice that our original integral, ∫exsinxdx, has emerged at the end of the second iteration. Believe it or not, this is exactly what we want. Let ∫exsinxdx = A. Substituting and carefully retracing the procedure, we arrive at
A = exsinx – [excosx + A]
or
A = exsinx – excosx – A
Adding A to both sides we get
2A = exsinx – excosx
Dividing both sides by 2, then resubstituting ∫exsinxdx for A, gives
A = ∫exsinxdx = (exsinx – excosx)/2 + C
I’ll be talking more about integral calculus in coming posts:)
Source:
Larson, Roland and Robert Hostetler. Calculus. Toronto: D C Heath and Company, 1989.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
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