Download Lecture 2: Techniques for Integration I. Integration by Parts, part II

Lecture 2: Techniques for Integration
I. Integration by Parts, part II
Z
Z
udv = uv −
vdu
Choose u and dv wisely:
Use the acronym IPET as a guideline for choosing u. Check the integrand in the order:
• I = inverse func. (ex. arctan(x), ln(x))
(one IBP at a time)
• P = polynomial (ex. 5x3)
• E = exponential func. (ex. ax, e3x)
(sometimes 2 IBPs)
• T = trigonometric func. (ex. cos(2x), tan(x))
(E & T interchangeable)
Z
ex.
ln xdx
(x ln x − x + c)
1
Z
ex. Evaluate
1
arctan xdx
0
√
(x arctan x + ln( 1 + x2 ) = π/4 − (1/2) ln 2)
2
Z
ex.
ex sin xdx
([ex (sin x − cos x)]/2 + c)
3
SUMMARY: COMMON INTEGRALS
USING INTEGRATION BY PARTS
1. For integrals of the form
Z
Z
Z
xnabx dx, xn sin ax dx, xn cos ax dx
Let u = xn and differentiate u all the way till
du = 0.
2. For integrals of the form
Z
Z
Z
xn ln x dx, xn arcsin ax dx, xn arctan ax dx
Let u = ln x, arcsin ax, or arctan ax and differentiate u just once and evaluate the ’left
over’ integral. (may need to repeat IBP)
3. For integrals of the form
Z
Z
eax sin cx dx, abx cos cx dx
Let u = abx and differentiate u twice and
evaluate the ’left over’ integral.
4
Now You Try It (NYTI):
Z
•
Z
•
Z
•
√
sin( x)dx
x5xdx
x5 ln xdx
√
√
√
(−2 x cos( x) + 2 sin( x) + c)
x
( x5
ln 5 −
6
5x
(ln 5)2
+ c)
( x6 ln x −
1 x6
6 6
5
+ c)
Z
•
t
ex sin(t − x)dx
( 21 (et − cos t − sin t))
0
Z
•
x ln(1 + x)dx
( 21 x2 ln |1 + x| − 14 x2 + 12 x −
1
2
ln |1 + x| + c)
6
Z
•
x sec2(2x)dx
( 12 [x tan(2x) +
x arctan xdx
( 21 x2 arctan x − 12 x +
1
2
ln | cos(2x)|] + C)
Z
•
1
2
arctan x + c)
Z
•
cos x ln(sin x)dx
(sin(x) ln | sin x| − sin x + c)
7