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Instead of using the formula $\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$, would $\int f'(x)g(x)dx = f(x)g(x) - \int f(x)g'(x)dx$ work with solving integration by parts problems? Based on the derivation of this function from the product rule in differentiation, it seems like it would work (since they are equivalent) but the numbers seem to get much uglier (at least, from the problems that I have attempted this secondary formula with).

I tried to solve $\int x\cdot \sin (2x)dx$ in this way, and easily obtained the answer using the regular formula, but the numbers become different as you solve it with the secondary formula.

Thank you!

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I think this should work. But you should differentiate your answer and check whether you get $x\sin{2x}$ –  user9413 Jul 3 '11 at 5:28

3 Answers 3

up vote 2 down vote accepted

There is no a priori preference to one or the other. Simply consider $\int {\sin (2x)x \,dx}$.

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However, for any specific example, you should check which is preferable. In your example of $\int {x\sin (2x)dx}$ it is obvious that the former should be used. –  Shai Covo Jul 3 '11 at 5:42

What the post calls two formulas is a single procedure. The real issue is not which "formula" to use. We have to decide the following: "How shall I express the thing I am trying to integrate as a product of two things, one of which I will differentiate, and the other of which I will integrate?" If some term went "down" (differentiate) and another went "up" (integrate), would things look better? Usually what superficially looks good is good. After doing a few dozen well chosen problems, one develops a feel for the structure.

Rules of thumb like the "LIATE" rule are superficially attractive, but in my opinion they cause long term damage.

Many students who are blessed with excellent memories succeed in school math by memorizing rules and examples. As a short term strategy it can work well. If the student needs to get through a math course and go on to do no more mathematics, the strategy may be useful.

But if one will continue to need to use mathematics, the strategy is in the long run disastrous. I have seen many students with high grades in school mathematics get into real difficulties later, because they did not really know the material. All they have is retained is a dimming memory of a jumble of tricks.

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Basically, this should work. But I follow the below rule, for choosing $u$ and $dv$ accordingly

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Thanks! Very useful rule. I appreciate the help. –  Arthur Collé Jul 3 '11 at 6:52

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