# Integration by substitution Try integrating

```Integration by substitution
Try integrating:
 x  5 dx
5
Hint: Try a guess and differentiate it.
x  5
6
6
c
That was easy to guess, but lets try it with substitution:
 x 5 dx
5
Let u  (x  5)
 u dx
5
(the dx is a problem)
du
u  x 5,
1,dx  du
dx
So,

So we have:
u6
(x  5)6 
 u du  6  c  6  c
5
That’s a simple one …. now let’s try a harder one ……..
Try this integral:
2
4
x(x
3)
dx

2
u  x  3 so,
du
 2x
dx

and
du
dx 
2x
du
 xu 2x

1
u4 du

2
4
5
Integrate and then substitute: u  c 

10


x2  3
5
10
c
minutes).
7
6
x  4
1.
x  4 dx

2.
3.
4.
5.
 6x3x
 x
2
2
3

x  5 dx
2
xcos(x
 3)dx

x

x e
2
3
3x
 8 dx
3
4
dx
7
2
c
 8
4
c
4
3
2
2x 3  5




9
sin x 2  3
2
c
c
x2 4 
e
c
3
 x 3 dx
1
5
0
u  x 3
Take the upper and lower values of x from the
integral and find the corresponding values of u.
x 1,u  2
x  0,u  3
Now re-write the integral in
terms of u, as you did before.
u6 2
3 u du  6 
3
2 5
 64   729  665
   

 6   6  6
Confirm this result by using your Graphics Calculator.
minutes).
2
x2
1.  xe dx
1
 
25.9
x cosx 2  1dx
0.0787

2.

3.

4
0
3
1
x
dx
2
x  1


0.805
One step beyond….
Try this one:-

x
dx
2
x  3
du
(too easy!)
u  x  3,
1,dx  du
dx

u x3
x
 2 du Something wrong? - That x, but …….
x  u 3
u

u 3
du 
2
u
1 3 
 u  u2 du
3
 ln u   c
u
3
 ln x  3 
c
x  3

```