Choosing Your Players
2. The LIATE or ILATE Mnemonic
Okay, so how do you decide which part of your integrand becomes 'u' and which becomes 'dv'? There's a handy mnemonic device called LIATE (or ILATE, depending on your professor's preference) that can help guide your decision. It stands for:
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Logarithmic functions (like ln(x))
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Inverse trigonometric functions (like arctan(x))
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Algebraic functions (like x2, x3, etc.)
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Trigonometric functions (like sin(x), cos(x))
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Exponential functions (like ex)
The idea is that you choose 'u' to be the function that appears earliest on this list. The rest of the integrand then becomes 'dv'. Why this order? Because functions higher on the list tend to get simpler when you differentiate them (which is what you'll be doing to find 'du'), while functions lower on the list tend to stay relatively the same or become only slightly more complex when you integrate them (which is what you'll be doing to find 'v').
Let's look at an example. Suppose you need to integrate x sin(x) dx. Here, you have an algebraic function (x) and a trigonometric function (sin(x)). According to LIATE, the algebraic function comes first, so you'd choose u = x and dv = sin(x) dx.
Don't treat LIATE as gospel, however. It's a guide, not a rigid rule. Sometimes, you might need to tweak your choice based on the specific integral you're dealing with. Experience and practice are your best teachers here. And remember, sometimes it takes a few tries to find the combination of U and V that will lead you to an easier integral.
The Differentiation and Integration Dance: Finding DU and V
3. The Math Tango: A Step-by-Step Guide
Once you've chosen your 'u' and 'dv', the next step is to find 'du' and 'v'. This involves a bit of differentiation and integration, hence the name "integration by parts."To find 'du', simply differentiate 'u' with respect to x. So, if u = x, then du = dx. Pretty straightforward, right?To find 'v', integrate 'dv' with respect to x. This is where your integration skills come into play. For example, if dv = sin(x) dx, then v = -cos(x). Don't forget the constant of integration (C) but generally we leave it out at this intermediate step of finding V.
Now that you have u, dv, du, and v, you're ready to plug them into the integration by parts formula: u dv = uv - v du. It's like assembling the pieces of a mathematical puzzle.
Let's continue our example from above: x sin(x) dx. We have u = x, dv = sin(x) dx, du = dx, and v = -cos(x). Plugging these into the formula, we get: x sin(x) dx = x (-cos(x)) - (-cos(x)) dx.
Simplify and solve: x sin(x) dx = -xcos(x) + cos(x) dx = -x cos(x) + sin(x) + C. And there you have it! You've successfully integrated by parts.
Putting It All Together: A Worked Example
4. Conquering the Integral: A Real-World Scenario
Let's tackle a slightly more complex example to solidify your understanding. Suppose you want to evaluate x2 ex dx.Using LIATE, we choose u = x2 (algebraic) and dv = ex dx (exponential). This means du = 2x dx and v = ex.
Applying the integration by parts formula, we get: x2 ex dx = x2 ex - ex 2x dx = x2 ex - 2x ex dx. Notice that we now have a new integral to solve: x ex dx. This is where the "by parts" part really shines.
We apply integration by parts again to this new integral. This time, u = x and dv = ex dx, so du = dx and v = ex. Then, x ex dx = x ex - ex dx = x ex - ex + C.
Substituting this back into our original equation, we get: x2 ex dx = x2 ex - 2(x ex - ex) + C = x2 ex - 2x ex + 2ex + C. Ta-da! You've conquered a double integration by parts problem. It's like leveling up in the calculus game!
