Integrate by Parts, the IS-ID Way

In calculus, we learn to integrate products like x \ln{x} and e^x \sin{x} using the technique integration by parts. We use the formula,

\displaystyle \int uv' \, dx = uv - \int u'v \, dx ,

and using LIATE (Log-InvTrig-Algebraic-Trig-Exp), choose the right-leaning term as v' (e.g. x) and the left-leaning term  as u (e.g. \ln{x}). We integrate and differentiate respectively to get  v=\displaystyle \frac{x^2}{2} and u'=\displaystyle \frac{1}{x} respectively, substitute into the formula, simplify the expression and carry on our calculations. Rather than confuse you with which expression to use as \displaystyle v' and \displaystyle u, I propose a simpler presentation.

Before we discover this presentation, I want to clarify that this is just a simpler presentation of the same technique. The concept is the same. The presentation simply makes more intuitive sense. Also, I might derive the formula in another post. Integration by parts is essentially what I call the “reverse product rule”, and I might elaborate more next time. For now, lets take a look at the formula one more time, but this time I highlighted in green  v and  v' and red  u and  u', and swapped the order of multiplying (doesn’t change the formula mathematically):

\displaystyle \int v' \, u dx = v u  - \displaystyle \int v \, u' dx .

Notice that v' \, gets Integrated and u remains the Same in the first term, v u. Notice that v' \, gets Integrated and u gets Differentiated in the second term, \displaystyle \int v \, u' dx . Thus, we get this idea of Integrate * Same – Integrate * Differentiate, or succinctly put,

I S  - \displaystyle \int I \, D dx .

As mentioned, this is essentially the same method, but using these letters makes it easier for us to choose which terms to use. Let’s integrate x \ln{x} as an example.

Between x and \ln{x}, the right-leaning term is x. Thus, we Integrate that to get  \displaystyle \frac{x^2}{2}. The left-leaning term is \ln{x}. Thus, we Differentiate that to get \displaystyle \frac{1}{x}.

Using IS-ID, we plug in the relevant letters to get

\displaystyle \int x \, \ln{x} dx = \displaystyle \frac{x^2}{2} \ln{x}  - \displaystyle \int \displaystyle \frac{x^2}{2} \displaystyle \frac{1}{x} dx  \displaystyle = \frac{x^2}{2}\ln{x} - \frac{1}{2}\int x \, dx .

Integrating the remaining portion, we get our final result,

\displaystyle \int x \ln{x}\, dx = \frac{x^2}{2}\ln{x} - \frac{x^2}{4} + C.

I hope this alternative presentation will help you more effectively and integrate by parts with better understanding. As an exercise to the reader, find \displaystyle \int e^x \sin{x}\, dx. The final answer, for your reference, is \displaystyle \frac{e^x}{2} (\sin{x} - \cos{x}) and you get bonus marks if you can explain how this answer can be improved.



Author: joelkindiak

Build people up. Point them to Christ.

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