Tutorials with examples and detailed solutions and exercises with answers on how to use the technique of integration by parts to find integrals.
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
Exercises: Use the table of integrals and the method of integration by parts to find the integrals below. [Note that you may need to use the method of integration by parts more than once].
1. x cos(x) dx
2. x e^{2x} dx
3. x^{1/3} ln(x) dx
4. ln(x) / (x^{2}) dx
5. x^{3} cos(x) dx
6. x^{2} e^{3x} dx
7. e^{x} cos(2x) dx
Answers to Above Exercises
1. x sin(x) + cos(x) + K
2. (x/2)e^{2x}  (1/4)e^{2x} + K
3. (3/4)x^{3/4} ln(x)  (9/16)x^{3/4} + K
4. ln(x) / x  1/x + K
5. 3(x^{2}  2)cos(x) + (x^{3}  6x)sin(x) + K
6. (9x^{2} + 6x + 2) e^{3x} dx + K
7. (1/5)(e^{x}cos(2x) + 2 e^{x}sin(2x))
More references on
integrals and their applications in calculus.
