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} + C

### 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) + C

### 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)) + C

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

### 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} + C

### 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

= 3 x sin(x/3) + 9 cos(x/3) + C

### 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 + C

### 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} + C

### 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)] - 4∫e^{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) ) + C

## 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) + C

2. (x/2)e^{2x} - (1/4)e^{2x} + C

3. (3/4)x^{3/4} ln(x) - (9/16)x^{3/4} + C

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

5. 3(x^{2} - 2) cos(x) + (x^{3} - 6 x) sin(x) + C

6. - ( 1/27 )( 9 x^{2} + 6 x + 2 ) e^{-3x} dx + C

7. (1/5) ( e^{x} cos(2x) + 2 e^{x} sin(2x) ) + C

## More References and links

integrals and their applications in calculus.