Unit 8 - Objective 6 - Integration by Parts

This section explains how the product rule for derivatives [d(uv) = u dv + v du] translates to integrals. We can rewrite the integral of a product as another integral which is (hopefully) easier to integrate than the original. Integrating the product rule gives:

d (uv) = u dv + v du    so that  uv =   u dv + v du

To use this formula, perform the following steps:

1. Identify (u) and (dv) in the given integral.
2. Find du (by taking the derivative of u) and v (by integrating dv, leaving off the constant, C, till later).
3. Substitute those expressions into the formula above.
4. Integrate the resulting integral.

For a definite integral, the above formula becomes:

Problems 1-3 below use integration by parts to find integrals which have "extra X's". If the X's were not there, we could have found the integral. In this case, pick u = "extra x".

Problem 1:

Problem 2:

7x sin 4x dx

The 7 factors out. Let u = x, dv = sin 4x dx, then du = 1 dx and v = sin 4x dx = -(1/4)cos 4x.

7x sin 4x dx	= 7 x sin 4x dx
	= 7 [u v - v du] = 7 [ x ((1/4)cos 4x) - ((1/4)cos 4x) dx]
	= 7 [(-1/4)x cos 4x + (1/4) cos 4x dx]
	= 7 [(-1/4)x cos 4x + (1/4) (1/4) cos 4x (4) dx]
	= 7 [(-1/4)x cos 4x + (1/16) sin 4x + ]
	= (-7/4) x cos 4x + (7/16) sin 4x + C

Problem 3:

	Rewrite using exponents

The problems below show how integration by parts can be used to integrate products where one function has an "algebraic" derivative.

Functions like ln x, x, x, have "algebraic" derivatives. Select u to be this kind of function.

Problem 4:

x³ ln x dx	let u = ln x, then dv = x³ dx and we get du = (1/x) dx,
(there should be a 1/4 in front of the integral on the 4th line from the bottom)

Problem 5:

	We have a formula for the derivative of an Arctangent, not the integral.
Let u = 3x, then dv = dx (all that's left) and du =

	=
	=
	=
	= x 3x - (3/18) ln | 1+ 9x2 | + C
	= x 3x - (1/6) ln (1+ 9x2) + C3