Integration by Parts

We describe how to perform integration by parts. This webpage also describes various trigonometry reduction properties and integrals involving quadratic polynomials.

The basic approach can be summarized as follows:

For definite integrals, integration by parts takes the following form:

Proof: By the derivative product rule (see Differentiation Techniques):

Examples

Example 1:

We repeat Substitution Rule Example 2 from Differentiation Techniques

We set u = sin² x and dv = sin x dx. Thus, du = 2 sin x cos x and v = – cos x. Hence

We note that ∫ sin³ x dx occurs on both sides of the equation. Solving for ∫ sin³ x dx, we obtain

and so

which is the same answer we got in Differentiation Techniques.

Example 2:

We use the substitution t = √x, and so dt = dx/(2√x) = dx/(2t). Hence

We now set u = t and dv = sin x dx. Thus, du = 2 sin x cos x and v = – cos x. Hence

Thus

Trig reduction properties

In each case, the exponent is reduced by two in each step. If n is odd then eventually you need to take the integral of a trig function, as described in Trig Integrals of Integration Techniques. If n is even, then eventually you need to take the integral ∫ dx = x + C.