5.8
Inverse Trigonometric
Functions: Integration
and Completing
the Square
du
u
=
arcsin
+
C
Ú a2 - u2
a
du
1
u
Ú a 2 + u 2 = a arctan a + C
u
du
1
Ú u u 2 - a 2 = a arc sec a + C
dx
x
= arcsin + C
2
2
4- x
Ex.
Ú
Ex.
dx
=
2
Ú 2 + 9x
1
=
3
Ú
( 2)
Ú
du
2
+ ( u)
( 2)
1
2
dx
2
+ ( 3x )
2
Let u = 3x
du = 3 dx
du
= dx
3
3x
=
arctan
+C
3 2
2
Ex. Integrate by substitution.
Ú
=Ú
=Ú
dx
e -1
2x
Ú
(e )
x 2
-1
1
du
2
(u ) - 1 u
du
u
=
dx
(u )
2
-1
= arc sec
u
1
Let u = ex
du = ex dx
du
= dx
x
e
du
= dx
u
+ C = arc sec
e
x
1
+C
Ex. Rewriting the integrand as the sum of two quotients.
x+2
Ú
4- x
2
Let u = 4 –
dx = Ú
x2
du = -2x dx
du
= dx
- 2x
= 2Ú
dx
Ú
x
4- x
2
dx + Ú
2
4- x
2
dx
x du
u
2u1 2
=Ú
du =
-2
-2
u (- 2 x )
x
= 2 arcsin
2
2
2
2 -x
-1 2
- 4- x
2
Final Answer
x
- 4 - x + 2 arcsin + C
2
2
Ex. Integrating an improper rational function.
3x - 2
Do long division and then rewrite the
Ú x 2 + 4 dx integrand as the sum of two quotients.
12 x + 2 ˆ
Ê
= Ú Á 3x - 2
˜dx
x +4 ¯
Ë
12 x
2dx
= Ú 3 xdx - Ú 2
dx - Ú 2
x +4
x +4
2x
dx
= Ú 3 xdx - 6 Ú 2
dx - 2 Ú 2
x +4
x +4
3
2
(
)
3x
x
2
=
- 6 ln x + 4 - arctan + C
2
2
Ex. Completing the Square
dx
=
Ú x2 - 4x + 7
=
=
Ú
Ú
Ú
dx
2
( x - 2) + 3
du
(u)
2
+
( 3)
dx
2
( x - 4 x + 4) + 7 - 4
Let u = x – 2
du = dx
2
1
x-2
=
arctan
+C
3
3
Ex. Completing the square when the leading
coefficient is not 1.
dx
Ú 2 x 2 - 8 x + 10
First, factor out a 1/2
Now complete the square
in the denominator.
1
dx
1
dx
= Ú 2
= Ú 2
2 x - 4x + 5 2 x - 4x +
5
1
dx
1
dx
= Ú 2
= Ú
2
2 x - 4 x + 4 + 5 - 4 2 ( x - 2) + 1
1
du
1
( x - 2)
Let u = x – 2
=
=
arctan
+
C
2
Ú
du = dx
2 (u ) + 1 2
1
Find the area of the region bounded by the graph of
f(x) =
1
3x - x
2
3
9
, the x-axis, and x = , and x = .
2
4
94
2
Ú
32
1
1
2
1
3x - x
2
dx
Factor out a neg. inside the rad.
94
Ú
32
94
=
Ú
32
94
1
3x - x
2
dx =
Ú
32
(
1
- x - 3x
2
94
1
9ˆ 9
Ê 2
- Á x - 3x + ˜ +
4¯ 4
Ë
dx =
Ú
32
dx
)
1
2
3ˆ
Ê3ˆ Ê
Á ˜ -Áx - ˜
2¯
Ë2¯ Ë
2
dx
È
x - (3 2 )˘
1
p
p
= Íarcsin
= arcsin - arcsin 0 = - 0 =
˙
3 2 ˚3 2
2
6
6
Î
94