Integration by Parts


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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 ex dx

Solution to Example 1:

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

3 x ex dx

= 3 x ex dx

= 3 ( x ex - 1 . ex dx)

= 3 x ex - 3 ex + 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

x2 cos(x) dx

Solution to Example 3:

Let u = x2 and dv/dx = cos(x), hence du/dx = 2x and v = sin(x)

x2 cos(x) dx

= x2 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

x2 cos(x) dx = x2 sin(x) - 2 (- x cos(x) + sin(x)) + K

= x2 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 = x2 / 2

x2 ln(x) dx

= (x2 / 2) ln(x) - (x2 / 2) (1 / x) dx

= (x2 / 2) ln(x) - (1/4) x2 + 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

x2 (ln(x))2 dx

Solution to Example 7:

Let dv/dx = x2 and u = (ln(x))2 and use integration by parts as follows

I = x2 (ln(x))2 dx

= [x3 / 3 (ln(x))2] - (x3 / 3)(2 ln(x) / x) dx

= [x3 / 3 (ln(x))2] - (2/3)x2 ln(x) dx

We now let dv/dx = x2 and u = ln(x) and use the integration by parts one more time

= [x3 / 3 (ln(x))2] - (2/3)[ (x3 / 3 ln(x) - (x3 / 3) (1/x) dx ]

= [x3 / 3 (ln(x))2] - (2/9)(x3 ln(x) + (2/9) x2 dx

Integrate the last term

= [(ln(x))2 x3 / 3 ] - (2/9)(x3 ln(x) + (2/27)x3 + K


Example 8: Use integration by parts to evaluate the integral

ex sin(2x) dx

Solution to Example 8:

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

I = ex sin(2x) dx

= [ex sin(2x)] - ex 2 cos(2x) dx

We now let V = ex and du/dx = ex and use the integration by parts one more time

= [ex sin(2x)] - 2[ex cos(2x)] + 2 ex 2 (- sin(2x)) dx

= [ex sin(2x)] - 2[ex cos(2x)] - 4ex sin(2x) dx

We can write the integral I as follows

I = [ex sin(2x)] - 2[ex cos(2x)] - 4I

and solve the last equation for I

I = (1/5)(ex sin(2x) - 2ex 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 e2x dx

3. x1/3 ln(x) dx

4. ln(x) / (x2) dx

5. x3 cos(x) dx

6. x2 e-3x dx

7. ex cos(2x) dx

Answers to Above Exercises

1. x sin(x) + cos(x) + K

2. (x/2)e2x - (1/4)e2x + K

3. (3/4)x3/4 ln(x) - (9/16)x3/4 + K

4. -ln(x) / x - 1/x + K

5. 3(x2 - 2)cos(x) + (x3 - 6x)sin(x) + K

6. -(9x2 + 6x + 2) e-3x dx + K

7. (1/5)(excos(2x) + 2 exsin(2x))

More references on integrals and their applications in calculus.



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Updated: 2 April 2013

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