Page 14 - DMTH202_BASIC_MATHEMATICS_II

Unit 1: Integration

If we let u = 1 – x , then du = – 2xdx, but we don't have –2xdx. Notes
2
Also, integration by parts will not function.
2 –3/2
2 –1/2
If we let u = (1 – x ) , then du = x(1 – x ) dx. dv = dx, and so v = x.
1 x x 2

Putting it together, we have:  2 dx  2  3/2 dx , which has only made the

1 x 1 x 1 x 2 

integral worse.
But what happens if we let x = sin. Observe that the quantity inside of the square root turns out
t be 1 – sin  = cos , by the Pythagorean Identity above.
2
2
Also, observe that dx = cosd. Unlike before, here we do not have to solve for dx; it is already
given to use.
So, now we just plug everything in and solve.

1 x cos cos
 dx  cos d   d   d   d    C

2
2
1 x 2 1 sin  cos  cos


The question remains, what is ? We need to solve for  in terms of x.
If we gaze above, we observe such a relationship. Recall, we had let x = sin. That means that
–1
 = sin (x) or arcsin(x).
1
So, we have that  dx  arcsin   x  . C

1 x 2
1
Example: Evaluate 2  dx .
x  4x  13
First we complete the square in the denominator by observing that 13 = 4 + 9 and then that
2
2
provides us x + 4x + 4 + 9 = (x + 2) .
1 1 1
2  dx   dx   dx .
2
2

x  4x  13 x  4x  4 9 x   2 3 2
Then we let x + 2 = 3tan. Observe that dx = 3sec d.
2
After substituting, we have
2
2
2
1 3sec  d 3sec  d 3sec  d 1 1
dx
 2 2  2   2   2   d    . C

x   2  3 3tan  9 9tan   9 9sec  3 3
 x  2 
Solving for , we have   arctan   .
 3 
1 1  x  2
Thus, 2  dx  arctan    C .
x 4x  13 3  3 
dx
Example: Evaluate 2 

d 4 x 2

LOVELY PROFESSIONAL UNIVERSITY 9