Download Math 104 – Calculus 8.4 Trigonometric Subs tu ons

Math 104 – Calculus 8.4 Trigonometric Subs=tu=ons Math 104 -­‐ Yu Trigonometric subs=tu=on •  Some=mes it is very helpful to use trigonometric iden==es to simplify involving radical expressions. •  It can be used to evaluate integrals containing the following expressions: 1.
p
a2
2.
p
a2 + x2 or
3.
p
x2
x2
1
a2 + x2
a2
•  Don’t forget to change the limits of the definite integrals! Math 104 -­‐ Yu 2 -­‐ x2 2-x2
Square r
oot o
f a
Integrals
with
square
root
of
a
Integrals with square root of a2-x2
= axsin
Let xLet
= a✓ sin ✓
p p
p p
2
Then Then
a2 ax22 =x2 =a2 a2a2 sin
a2 sin✓2 ✓
q q
2 2
2 (1
2 (1
=
a
sin
=
a
sin
✓) ✓)
p
p
2✓
2 cos
=
a2 2cos
=
a
✓
a| ✓|
cos ✓|
= a|=cos
x
x
We✓have
✓ = arcsin
We have
= arcsin
a
a
a

Since Since
ax
a, x  a
⇡
⇡

✓

Then
⇡
⇡
2
2 and cos ✓ > 0
Then
✓  and cos ✓ 0.
2 p
2
Thus
p
a2 x2 = a cos ✓
Thus a2 x2 = a cos ✓.
Nicolas Fraiman
Math 104
Math 104 -­‐ Yu Nicolas Fraiman
Evaluate
1.
Z
p
1
Examples
Example 2
dx
p
x2 4 x2
Math 104 -­‐ Yu Nicolas
Fraiman
Math 104
Example 2. Find the area of the unit disk. Math 104 -­‐ Yu Tips To evaluate a definite integral we either undo the change of
variable, or find the limits in the new variable θ.
To undo the change of variable we use the reference
triangle to find formulas for functions of θ.
To find the new limits we use inverse trig. functions.
Nicolas Fraim
Math 104
Math 104 -­‐ Yu 2+x2
Integrals
with
square
root
of
a
2 + x2 2
2
Square r
oot o
f a
Integrals
with
square
root
of
a
+x
Let x = a tan ✓
Let x = a tan ✓
p
p
p
2 + x2 = p
2 + a2 tan2 ✓
a
a
Then
2
Then
a2 + x2 =q a2 + a2 tan ✓
q2
2
=
a
(1
+
tan
2
= a (1 + tan2 ✓)
✓)
pp
2✓
== aa22sec
sec2 ✓
==a|a|sec
sec✓|✓|
x
We have ✓ = arctan x
We have
✓ = arctan a
a
Since x can be any real
number,
⇡
⇡
Then
and sec
sec ✓✓ > 1.
Then
⇡2 2<✓ ✓<⇡22 and
0
p
p
2 + x2 = a sec ✓.
a
Thus
2
a + x2 = a sec ✓
Thus
Nicolas
Math 1Fraiman
04 -­‐ Yu Nicolas
Fraiman
Math 104
Examples
Example 3.
Find
Z
1
dx
2
1+x
!
!
!
4. Evaluate
Z
1
0
dx
(x2 + 4)3/2
Nicolas Fraiman
Math 104
Math 104 -­‐ Yu 2 -­‐ of
2 x2-a2
Integrals
with
square
root
Square r
oot o
f x
a
2-a2
Integrals
with
square
root
of
x
Let
x = a sec ✓
Let
x = a sec ✓
p
p
2
2 =
2 sec2 ✓
2
p
Then
xp
a
a
a
Then
x2 a2 p
= a2 sec2 ✓ a2
2 (sec2 ✓
= ap
1)
2
2
= a (sec ✓ 1)
p
p
2 tan2 ✓
=
a
2 tan2 ✓
=
a
= a| tan ✓|
= a| tan ✓|
x
x
We have
✓ = arcsec
x
1
We We
havehave
✓ = sec
,
note
that |x| a
✓ = arcsec
a
a
a
⇡
0  0✓✓ 2< ⇡ ,
If x
1.If Ifxax then
a
then
0✓ 2
a
then
p
so
p
2 = a tan ✓.
so
x2px2 2a2a=
✓ ✓
x
a2a=tan
a tan
⇡
⇡  ⇡

✓
⇡
2.
If
x

a
then
x

a
If If px  then
2
2<✓ ✓⇡⇡,
a then
2
2
2 =
x
a
a
tan ✓.
so p
sox2p 2a2 =2 a tan
✓ ✓
x
a = a tan
Nicolas
Fraiman
Math 104 -­‐ Yu Nicolas
Math 104 Fraiman
5. Evaluate
Z
p
dx
.
2
9x
4
Example Math 104 -­‐ Yu Summary 1. a 2 − x 2 Let x = a sin θ
a2 + x2
2.
Math 104 – Rimmer
8.3 Trig. Substitution
Let x = a tan θ
3.
x2 − a2
Let x = a sec θ
2
a 2 + x 2 = a 2 + ( a tan θ )
= a 2 − a 2 sin 2 θ
= a 2 + a 2 tan 2 θ
= a 2 sec2 θ − a 2
= a 2 (1 − sin 2 θ )
= a 2 (1 + tan 2 θ )
= a 2 ( sec 2 θ − 1)
= a 2 cos 2 θ
= a 2 sec2 θ
= a 2 tan 2 θ
a 2 − x 2 = a 2 − ( a sin θ )
a 2 − x 2 = a cos θ
x
x = a sin θ ⇒ = sin θ
a
−π
π
≤θ ≤
2
2
a 2 + x 2 = a sec θ
x
x = a tan θ ⇒ = tan θ
a
⇡
−π
π ⇡
22
≤<
θ ≤✓
<
2
2
a
2
a +x
θ
2
a −x
2
a
( a sec θ )
2
− a2
x 2 − a 2 = a tan θ for xx > aa
x 2 − a 2 = − a tan θ for x
x <−a a
x
x = a sec θ ⇒ = sec θ
a
π
π
x
θ
x2 − a2 =
0 ≤θ <
2
x
2
2
2
<θ ≤π
x
x2 − a2
θ
a
Math 104 -­‐ Yu Completing squares
Evaluate
6.
Z
Comple=ng squares 2
1
dx
p
4x x2
!
!
Hint: Write 4x
x2 = ( 4 + 4x
=4
(x
x2 ) + 4
2)2
Nicolas Fraiman
Math 104
Math 104 -­‐ Yu Remember u-substitution
!
Remember u-­‐subsitu=on ! •
Evaluate
Z
p
x
x2 + 4
dx
!
!
•
Don’t use a trig. substitution if a u-substitution is easier.
Nicolas Fraim
Math 1Math
04 -­‐104
Yu