5.7 Inverse Trig Functions: Integration 1. Integrate functions whose antiderivatives involve inverse trig functions 2. Use the method of completing the squares to make an integration fit. 3. Review basic integration rules involving elementary functions.
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5.7 Inverse Trig Functions: Integration
1. Integrate functions whose antiderivatives involve inverse trig functions
2. Use the method of completing the squares to make an integration fit.
3. Review basic integration rules involving elementary functions.
Integrals Involving Inverse Trigonometric Functions.The derivatives of the Inverse trig functions fall into three pairs, where one is just the negative of its paired function.Arcsin and arccos pair up, arctan and arccot pair up, and arcsec and arccsc pair up. The main three rules are below.
Ex 1 p. 380 Integration with Inverse Trig Functions
2.
9 4
dxa
x
To make the form fit, a = 3 and u = 2x. Then du = 2dx
2
1 2
2 9 4
dx
x
This looks like an arcsin form
1 2arcsin
2 3
xC
2.
6 9
dxb
x 6, 3 , 3a u x du dx
2 2
1 3=
3 6 3
dx
x1 3
arctan3 6 6
xC
Ex 1 continued
2.
4 9
dxc
x x a = 3, u = 2x, du = 2dx
2 2
2=
2 2 3
dx
x x
21arcsec
3 3
xC
Ex 2 p.381 Integration by Substitution
2
1x
dxFind
e 2 1
xdue
u
let . Then or x x
x
duu e du e dx dx
e
2 1
duu
u
2 1
du
u u
Now it fits the arc sec
rule
1sec
1 1
xearc C sec xarc e C
Ex 3 p.381 Rewriting the sum of two quotients
2
2Find
4
xdx
x
2 2
2
4 4
xdx dx
x x
Doesn’t really fit
one of the rules as is.Rewrite to use power rule and arcsin rule
1/ 22
2
12 4 2
2 4
dxx x dx
x
12 241
2arcsin12 2
2
x xC
24 2arcsin2
xx C
Completing the Square:Helps when quadratic function is involved in the integrand. Just to remind you of the process, start with lead coefficient of 1. 2 2 2 2
2 2
2 2 2
b b b bx bx c x bx c x c
x
Ex 4 p. 382 Completing the square
2Find
4 7
dx
x x Looking through our possible rules, we see arctan has sum of two squares in denominator.
2 24 7 ( 4 4) 4 7x x x x 2( 2) 3x
2( 2) 3
dx
x
3, 2,a u x du dx
1 2arctan
3 3
xC
If the lead coefficient is not one, factor it out before completing the square. This also works if it is a negative lead coefficient.
22 8 10x x 22( 4 5)x x 22[( 4 4) 4 5]x x 22[( 2) 1]x
23x x 21( 3 )x x 2 2
3 3
2 2x
2 22 3 3
( 3 )2 2
x x
2 23 3
2 2x
Ex 6 p. 382 Completing the square (negative lead coefficient)
2
1 3 9Find the area of the region bounded by ( ) , the x-axis, the lines x= and x=
2 43f x
x x
Solution:
94
23
23
dx
x x
94
2 23
2 3 32 2
dx
x
3 3, ,2 2a u x du dx
94
32
32arcsin
32
x
1arcsin arcsin 0
2
06
0.5235987756
Notice #4,5 not in front of book
Ex 6 p. 382 Comparing integration problems.
Looking at derivative rules, you now have a way to differentiate any elementary function. With integration this is far from true. We have to make things fit the right form or we can’t integrate.
2
dx.
x x 1a
2
x dx.
x 1b
2
dx.
x 1c
2
dx.
x 1a
x Fits arcsec rule
1sec
1 1
xarc C secarc x C
2
x dx.
x 1b
Fits power
rule
12 21
2 12
x x dx
2 1x C
12 21112
2
xC
2
dx.
x 1c
Doesn’t fit any rule so far – see p. A22, #31
Ex 7 p. 384 Comparing Integration problems
dx.
lna
x xln dx
. x
bx . ln c x dx
. ln
dxa
x xLet u = lnx then du=(1/x)dxdu
u ln ln x C
Let u = lnx then du = (1/x)dx
ln .
x dxb
x udu1
1
nuC
n
2(ln )
2
xC
. ln dx c x We don’t have a rule for this – p.A27 #87
Notice that in both Ex 6 and Ex 7, the simplest ones are the hardest to find rules for.
5.7 p. 385/ 1-45 every other odd, 53-54, 63-69 odd, 75-77, 83
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