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 + 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
∫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)) + C
= x2 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 = x2 / 2
∫x2 ln(x) dx
= (x2 / 2) ln(x) - ∫ (x2 / 2) (1 / x) dx
= (x2 / 2) ln(x) - (1/4) x2 + 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
∫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 + C
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)] - 4∫ex 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) ) + 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 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) + C
2. (x/2)e2x - (1/4)e2x + C
3. (3/4)x3/4 ln(x) - (9/16)x3/4 + C
4. -ln(x) / x - 1/x + C
5. 3(x2 - 2) cos(x) + (x3 - 6 x) sin(x) + C
6. - ( 1/27 )( 9 x2 + 6 x + 2 ) e-3x dx + C
7. (1/5) ( ex cos(2x) + 2 ex sin(2x) ) + C
More References and links
integrals and their applications in calculus.