I will show you an EASIER WAY to Integrate by Parts

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1

<p>I don't know how to use math notations on here, so bear with me. This is an easier method than the traditional way. </p>

<p>Let SS be the integral sign:</p>

<p>EXAMPLE #1:</p>

<p>SS (x^2)sinx dx</p>

<p>let u = x^2
let v' = sinx</p>

<p>u: x^2, 2x, 2, 0 (I kept taking the derivative)</p>

<p>v': sinx, -cosx, -sinx, cosx (I kept taking the integral)</p>

<p>So now the answer is:</p>

<p>SS (x^2)sinx dx = -x^2cosx+2xsinx+2cosx+C</p>

<p>EXAMPLE #2:</p>

<p>SS lnx dx</p>

<p>let u = lnx
let v' = 1 (since there are nothing else to set as v', we will always pick 1)</p>

<p>u: lnx, 1/x</p>

<p>v': 1, x</p>

<p>So now: SS lnx dx = xlnx - SS 1 dx = xlnx - x + C</p>

<p>EXAMPLE #3:</p>

<p>SS e^2xcos3x dx</p>

<p>let u = e^2x
let v' = cos3x</p>

<p>u: e^2x, -(2e^2x), 4e^2x</p>

<p>v': cos3x, 1/3(sin3x), -1/9(cos3x)</p>

<p>So now: SS e^2xcos3x dx = 1/3e^2xsin3x+2/9e^2xcos3x - 4/9 SSe^2xcos3x dx</p>

<pre><code> = 13/9 SS e^2xcos3x dx = e^2x/9(3sinx+2cos3x)
</code></pre>

<p>FINAL ANSWER : SS e^2xcos3x dx = e^2x/13(3sin3x+2cos3x)+C</p>

<p>Any questions, just let me know ;)</p>

2

<p>This is a special case and is called the Tabular method…</p>

3

<p>Question: Can you say what the partial derivative your method is?
Comment: Learn latex and post the code as plain text. That way at least a few more people can understand it…</p>

4

<p>So basically the same method as everyone uses, except you iterate it.
Faster perhaps, though I’d say ultimately useless because most computations such as these are best done by a computer. The concept is more important than the method after a while.</p>

5

<p>Yes, exactly the same method except you make a table and iterate it. It’s just faster than the old way.</p>

6

<p>Ah, I see what you’re doing now. Well, Mr. Styrkur, you must be wanting a lot of students to fail their calculus tests really hard, as you forgot to mention this algorithm only works when u is a polynomial.</p>

7

<p>

Two of these examples were non-polynomial.
The problem is, doing it like this doesn’t really teach you anything. Past a Calc II course, this really doesn’t give you much.</p>

8

<p>Forgot to mention, on the tables, the sign oscillates from +/-. So the first table was a postive and the second was a negative… so on.</p>

<p>More explanation:</p>

<p>For the first example, I stopped until the x^2 turned to zero when you keep taking the derivative.</p>

<p>Example 2: I stopped when the product across can be easily integrated: 1/x and x then becomes the integrand.</p>

<p>Example 3: I stopped as soon as it looks like the original function… 4e^2x and -1/9(cos3x) are very similar to the original: e^2xcos3x</p>

<p>This takes practice.</p>

<p>ALSO, this is the only technique I use for Integration By Parts. I always make that table.</p>

9

<p>This is no differen than normal integration by parts, you just aren’t writing intermediate steps. The problem is tht if you don’t already understand integration by parts, you won’t know when to stop your iterating anyway except when the problem naturally provides you with a stopping point such as in the case of u being a polynomial.</p>