Integration By Parts

Here is a guide on Integration by Parts. It is a tricky Calculus topic at first but it gets easier with practice.
Before continuing, one should be familiar with antiderivatives, the product rule and substitutions with integrals.


Topics


What is Integration by Parts?

Integration by Parts is an integration method for integrating functions like this:

\displaystyle \int x \cdot \text{e}^{x}\, dx
Here x is a polynomial and \text{e}^{x} is an exponential function. Another example is:

\displaystyle \int \sin(x) \cdot \text{ln}(x) \, dx
The natural question is how do I integrate these things!? We have to use a technique called Integration By Parts.


Intergration by Parts: The Formula

The formula for Integration by Parts is:

(1) \displaystyle \int u \, dv = u\, v - \int v \, du

One could ask what are u, v, du, and dv? We will look at the derivation of the formula.

To start, the product rule gives us:

\displaystyle (f(x) \, g(x))' = f(x) \, g'(x) + f'(x) \, g(x)

Integrating both sides gives us:

\displaystyle \int (f(x) \, g(x))' dx = \int f(x) \, g'(x) dx + \int f'(x) \, g(x) dx

By the Fundamental Theorem of Calculus, the integral of a derivative is the function (integrand) itself. The left side is just (f(x) \, g(x)) as follows:

\displaystyle f(x) \, g(x) = \int f(x) \, g'(x) dx + \int f'(x) \, g(x) dx

The substitutions u = f(x), v = g(x) along with du = f'(x) \, dx and dv = g'(x) \, dx turn the above line into:

(2) \displaystyle u \, v = \int u \, dv + \int v \, du

Rearranging the terms in (2) would give the integration by parts formula as given in (1) above.


The LIATE Memory Aid for Integration by Parts

You now know what u, v, du, and dv are. A natural question would be how do I know which function should be u and dv in the substitution for Integration by Parts? The LIATE principle can help determine what to pick for u and dv. The acronym LIATE stands for:

 

\displaystyle \text{L: Logarithmic Functions}
\displaystyle \text{I: Inverse Trigonometric Functions}
\displaystyle \text{A: Algebraic (or Polynomial such as } x^2)
\displaystyle \text{T: Trigonometric Functions}
\displaystyle \text{E: Exponential (}e^x \text{ for example)}

Top choices for u start from the letter L and go down and the top choices for dv start from the letter E for exponential and go up.

The rationale behind the LIATE principle is that logarithms have no known antiderivative so they are a common choice for the u substitution and that the antiderivative of an exponential such as e^x is an exponential.

An alternate acronym is LIPTE where the only difference is that the A for algebraic turns into P for polynomial.


Some Examples of Integration by Parts with LIATE

Example One

From before we had:

\displaystyle \int x \cdot \text{e}^{x} \text{ d}x

The algebraic function is x and \text{e}^{x} is the exponential function. Our choice for u is x (algebraic is higher than exponential) and dv would be \text{e}^{x} \, dx}. We would then have v = \text{e}^{x} and du = 1 \cdot dx.

Substituting the components u, du, v, and dv into the Integration By Parts formula \displaystyle \int u \, dv = u\, v - \int v \, du gives us:

Source: http://quicklatex.com/cache3/86/ql_72b65e5e9d6688cc2c58c1d6939f4c86_l3.png

Example Two

\displaystyle \int \sin(x) \cdot \text{ln}(x) \text{ d}x

The logarithm \text{ln}(x) would be the choice for u and we would have dv = \sin(x) \, dx. The derivative of \text{ln}(x) is \dfrac{1}{x} \, dx and the integral of \sin(x) \, dx is \, -\cos(x).

Example Three

\displaystyle \int \arctan(x) \cdot x \text{ d}x

We have an inverse trigonometry function such as \arctan(x) = \tan^{-1}(x) and an algebraic/polynomial function such as x. The choice for u is \arctan(x) and we would have dv = x \, dx. The other components would be du = \dfrac{1}{1 + x^2} \, dx and v = \dfrac{x^2}{2}.


Notes

The examples above were simple cases. Do be aware that product rule, quotient rule, and chain rule may be needed for determining du from u.

Multiple integration by parts may be needed at times.

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