Review:

The method of integration by parts may be used to easily integrate products of functions. If u and v are both functions of x then

u (dv/dx) dx = uv - (du/dx) v dx

In what follows K is a constant of integration.

__Example 1:__ Evaluate the integral

3 x e^{x} dx
__Solution to Example 1:__

Let u = x and dv/dx = e^{x}, hence du/dx = 1 and v = e^{x} and using the method of integration by parts, we obtain

3 x e^{x} dx

= 3 x e^{x} dx

= 3 ( x e^{x} - 1^{ . }e^{x} dx)

= 3 x e^{x} - 3 e^{x} + K

__Example 2:__ Evaluate the integral

x sin(x) dx
__Solution to Example 2:__

Let u = x and dv/dx = sin(x), hence du/dx = 1 and v = - cos(x)

x sin(x) dx

= x (-cos(x)) - 1 ^{.} (-cos(x)) dx

= - x cos(x) + sin(x) + K

__Example 3:__ Evaluate the integral

x^{2} cos(x) dx
__Solution to Example 3:__

Let u = x^{2} and dv/dx = cos(x), hence du/dx = 2x and v = sin(x)

x^{2} cos(x) dx

= x^{2} sin(x) - 2x sin(x) dx

We now need to apply the method of integration by parts to the integral 2x sin(x) dx
to obtain the final integral. In example 2 we determined the integral x sin(x) dx and we may use that result. Hence

x^{2} cos(x) dx
= x^{2} sin(x) - 2 (- x cos(x) + sin(x)) + K

= x^{2} sin(x) + 2xcos(x) - 2sin(x) + K

__Example 4:__ Evaluate the integral

x ln(x) dx
__Solution to Example 4:__

Let u = ln(x) and dv/dx = x, hence du/dx = 1/x and v = x^{2} / 2

x^{2} ln(x) dx

= (x^{2} / 2) ln(x) - (x^{2} / 2) (1 / x) dx

= (x^{2} / 2) ln(x) - (1/4) x^{2} + K

__Example 5:__ Evaluate the integral

x cos(x/3) dx
__Solution to Example 5:__

Let u = x and dv/dx = cos(x/3), hence du/dx = 1 and v = 3 sin(x/3)^{2} / 2

x cos(x/3) dx

= x (3sin(x/3)) - 1 ^{.} 3 sin(x/3) dx

= 3xsin(x/3) + 9 cos(x/3) + K

__Example 6:__ Use integration by parts to evaluate the integral

ln(x) dx
__Solution to Example 6:__

We first rewrite ln(x) as 1^{ . } ln(x), hence

ln(x) dx

= 1 ^{.} ln(x) dx

Let u = ln(x) and dv/dx = 1, hence du/dx = 1/x and v = x. Using integration by parts, we obtain

1 ^{.} ln(x) dx

= x ln(x) - x (1/x) dx

= x ln(x) - x + K

__Example 7:__ Use integration by parts to evaluate the integral

x^{2} (ln(x))^{2} dx
__Solution to Example 7:__

Let dv/dx = x^{2} and u = (ln(x))^{2} and use integration by parts as follows

I = x^{2} (ln(x))^{2} dx

= [x^{3} / 3 (ln(x))^{2}] - (x^{3} / 3)(2 ln(x) / x) dx

= [x^{3} / 3 (ln(x))^{2}] - (2/3)x^{2} ln(x) dx

We now let dv/dx = x^{2} and u = ln(x) and use the integration by parts one more time

= [x^{3} / 3 (ln(x))^{2}] - (2/3)[ (x^{3} / 3 ln(x) - (x^{3} / 3) (1/x) dx ]

= [x^{3} / 3 (ln(x))^{2}] - (2/9)(x^{3} ln(x) + (2/9) x^{2} dx

Integrate the last term

= [(ln(x))^{2} x^{3} / 3 ] - (2/9)(x^{3} ln(x) + (2/27)x^{3} + K

__Example 8:__ Use integration by parts to evaluate the integral

e^{x} sin(2x) dx
__Solution to Example 8:__

We first define I = e^{x} sin(2x) dx, v = sin(2x) and du/dx = e^{x} and use integration by parts as follows

I = e^{x} sin(2x) dx

= [e^{x} sin(2x)] - e^{x} 2 cos(2x) dx

We now let V = e^{x} and du/dx = e^{x} and use the integration by parts one more time

= [e^{x} sin(2x)] - 2[e^{x} cos(2x)] + 2 e^{x} 2 (- sin(2x)) dx

= [e^{x} sin(2x)] - 2[e^{x} cos(2x)] - 4e^{x} sin(2x) dx

We can write the integral I as follows

I = [e^{x} sin(2x)] - 2[e^{x} cos(2x)] - 4I

and solve the last equation for I

I = (1/5)(e^{x} sin(2x) - 2e^{x} cos(2x)) + K