2-9 - Last Week: Two Techniques of Integration Last week we...

Last Week: Two Techniques of Integration
Last week we talked about two tools for integrating difficult
functions:
I
Integration by Substitution (Change of Variable)
Z
x = b
x = a
f ( u ( x )) dx =
Z
u = u ( b )
u = u ( a )
f ( u )
dx
du
du
=
Z
u = u ( b )
u = u ( a )
f ( u ) x
0
( u ) du
=
Z
u = u ( b )
u = u ( a )
f ( u )
u 0 ( x )
du
I
Integration by Parts
Z
udv = uv -
Z
vdu
Z
b
a
udv = uv
b
a
-
Z
b
a
vdu

Remember the following “tips” we talked about:
I
If you see a compound function, think about trying a
substitution. Eg.,
Z
2
1
e
3 x
dx =
Z
6
3
e
u
1
3
du =
1
3
Z
6
3
e
u
du
I
If you’re integrating the product of a polynomial times a
function that anti-differentiates nicely (like e
x
or sin( x ) or
cos( x )), try integration by parts. Eg.,
Z
xe
x
dx = xe
x
-
Z
e
x
dx = xe
x
- e
x
+ C
I
If you have a polynomial times ln( x ), ignore the last tip—
differentiate the ln( x ) and anti-differentiate the polynomial.
Eg.,
Z
x
2
ln( x ) =
x
3
3
ln( x ) -
Z
x
3
3
1
x
dx
=
1
3
x
3
ln( x ) -
1
Z
x
2
dx
3

More Tips
We saw some instances where it was not obvious which tool(s) to
use.
I
Trigonometric Identities: Eg.,
I
Z
tan( x ) dx =
Z
sin( x )
cos( x )
dx
I
Z
sin
2
( x ) dx =
Z
1
2
-
cos(2 x )
2
dx
I
Z
sin
3
( x ) dx =
Z
(1 - cos
2
( x )) sin( x ) dx
which lead to a natural substitution.

More Tips
I
The “ · 1” Trick , eg.
Z
tan
- 1
( x ) dx =
Z
tan
- 1
( x ) · 1 dx
= x tan
- 1
( x ) -
Z
x ·
1
1 + x 2
dx
by integration by parts, and the integral
R
x
1+ x
2
dx on the right
can be solved with the substitution u = 1 + x
2
.
I
Cyclic Integration by Parts, eg.
Z
sin( x ) cos( x ) dx = sin
2
( x ) -
Z
cos( x ) sin( x ) dx
2
Z
sin( x ) cos( x ) dx = sin
2
( x ) + C
Z
sin( x ) cos( x ) dx =
1
2
sin
2
( x ) + C

More Techniques
0
Substitutions will not help you here! (At least, not directly.)
This integral requires the method of “partial fractions”.

Trigonometric Substitutions
The name of the game is the following trigonometric identities:
I
1 - sin
2
( θ ) = cos
2
( θ )
I
1 + tan
2
( θ ) =
cos
2
( θ )
cos 2 ( θ )
+
sin
2
( θ )
cos 2 ( θ )
= sec
2
( θ )
I
sec
2
( θ ) - 1 = tan
2
( θ )
Notice that in each case, the right hand side of the identity is a
perfect square.

Trigonometric Substitution—
R
1
1+ x
2
dx
Remember the identity
1 + tan
2
( θ ) = sec
2
( θ )
Maybe we could use the substitution x = tan( θ ) to turn the
denominator 1 + x
2
into sec
2
( θ )? This will turn out to be useful
because dx = sec
2
( θ ) d θ . Watch:
Z
1
1 + x
2
dx =
Z
1
1 + tan
2
( θ )
· sec
2
( θ ) d θ
=
Z
1
sec 2 ( θ )
sec
2
( θ ) d θ
=
Z
d θ
= θ + C = tan
- 1
( x ) + C

Trigonometric Substitution—
R
1
0
√
1 - x 2 dx
We already met this integral, and we know it gives the area of
one-quarter of the unit circle. This time, remember the identity
1 - sin
2
( θ ) = cos
2
( θ )
q
1 - sin
2
( θ ) = | cos( θ ) |
So perhaps it makes sense to use the change of variables
x = sin( θ ) dx = cos( θ ) d θ
This gives
Z
x =1
x =0
p
1 - x
2
dx =
Z
θ = π/ 2
θ =0
q
1 - sin
2
( θ ) · (cos( θ ) d θ )
=
Z
θ = π/ 2
θ =0
| cos( θ ) | · cos( θ ) d θ
=
Z
θ = π/ 2
θ =0
cos
2
( θ ) d θ
since cos( θ ) ≥ 0 for 0 ≤ θ ≤ π/ 2.

Trigonometric Substitution—
R
1
0
√
1 - x 2 dx
So a trigonometric substitution x = sin( θ ) shows that
Z
x =1
x =0
p
1 - x 2