Math 152, c Benjamin Aurispa
8.1 Integration by Parts
Recall the Product Rule. If u and v are functions of x , using differential notation, we have:
( uv )
0
= u v
0
+ v u
0
( uv )
0
= u dv + v du
Integrating both sides with respect to x, we would get: uv =
Z
[ u dv + v du ] =
Z u dv +
Z v du
Rearranging this, gives us the following:
Z u dv = uv −
Z v du
Integrals of the general form
Z x m e x dx
•
Z
1
0 xe
3 x dx
•
Z
( x
2
+ 5) e
− 2 x dx
1

Math 152, c Benjamin Aurispa
•
Z x
7 e
− x
4 dx
Integrals of the general form
Z x m cos x dx or
Z x m sin x dx
•
Z
3 x sin 2 x dx
•
Z
2 x
5 cos( x
2
) dx
2

Math 152, c Benjamin Aurispa
Integrals of the general form
Z x m ln x dx
•
Z x
7 ln x dx
•
Z ln x x 2 dx
•
Z ln x dx
3

Math 152, c Benjamin Aurispa
Other types: Inverse Trig and Loops
•
Z arcsin( x ) dx
•
Z e x cos 2 x dx
Another type of integral that would require a loop would be
Z cos(3 x ) sin(2 x ) dx
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Math 152, c Benjamin Aurispa
8.2 Trig Integrals
MUY IMPORTANTE identities for this section: sin
2 x + cos
2 x = 1 which means that sin
2 x = 1 − cos
2 x and cos
2 x = 1 − sin
2 x tan
2 x + 1 = sec
2 x which means that tan
2 x = sec
2 x − 1 cot
2 x + 1 = csc
2 x which means that cot
2 x = csc
2 x − 1 cos
2 x =
1
2
(1 + cos 2 x ) sin
2 x =
1
2
(1 − cos 2 x )
Goal of Trig Integrals: Factor out the DERIVATIVE of one of the trig functions. Whatever is left should be able to be rewritten using trig identities. Most of the time this is achieved by making sure whatever is left involve even powers. Then, make the appropriate u -substitution.
•
Z sin
3 x cos
6 x dx
•
Z sin
6 x cos
5 x dx
•
Z cos
3 x dx
5

Math 152, c Benjamin Aurispa
•
Z sin(5 x ) tan
2
(5 x ) dx
•
Z sin
7 x cos
3 x dx
•
Z sin
2 x dx
•
Z sin
2 x cos
2 x dx
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Math 152, c Benjamin Aurispa
•
Z tan
2 x dx
•
Z tan
8 x sec
4 x dx
•
Z x tan
5
( x
2
) sec
3
( x
2
) dx
7

Math 152, c Benjamin Aurispa
•
Z sec
4 x dx
•
Z cot
3 x csc x dx
•
Z sec x dx
8