Lesson Integration by parts (Calculus)

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About richard1234: BS in mathematics w/ computer science. Previous tutoring experience, and I have participated in the USAMO.

In integral calculus, integration by parts is a technique used to evaluate the integral of a complex expression, typically a product of two functions that do not work well together (e.g. a polynomial function with a trigonometric function). Here, we cover the basics of integration by parts, with an example.

Recall that the product rule says that, for two functions u and v (both in terms of x),

*[tex \LARGE \frac{d}{dx} (uv) = \frac{du}{dx}v + u\frac{dv}{dx}]

We can integrate both sides to obtain

*[tex \LARGE uv = \int v\, du + \int u\, dv], which implies that

*[tex \LARGE \int u\, dv = uv - \int v\, du]. This is the integration by parts formula. Note that u, v, du, and dv are in terms of x and dx.

Example
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Integrate:
*[tex \LARGE  \int x sin(x)\, dx].

We want to use a series of substitutions so that one "piece" of the integral represents a function u, and the other "piece" represents the derivative of another function, hence dv. A general rule of thumb is to let u be an expression that is easy to differentiate, and results in a simpler expression. Also, you want v to be an expression that is easy to integrate.

Suppose we let *[tex u = x]. Then, we can let *[tex dv = sin(x)dx]. From this, we can find du (the derivative of u), as well as v (by integrating dv).

*[tex du = dx]
*[tex v = -cos(x)]

Applying the integration by parts formula and substituting,

*[tex \LARGE \int u\, dv = uv - \int v\, du]

*[tex \LARGE \int x sin(x)\, dx = x(-cos x) - \int -cos(x)\,dx]

Here, we get the integral of -cos(x) dx, which is basic:

*[tex \LARGE \int x sin(x)\, dx = -x cos(x) + sin (x) + C], where C is an arbitrary constant.