Lenarz - Math 262
Strategy for Integration
Strategy for integration
1. Look to see if the integrand is a standard derivative that you already know.
2. Simplify the integrand if possible. Expand, factor or use trig identities to see if a
method jumps out.
3. Look for a substitution (including rationalizing substitutions).
4. Use the form of the integral to decide what method to use
(a) Products of powers of sin & cos (sec & tan or csc & cot)
(b) Rational functions
i. divide if improper
ii. factor denominator if possible, then try partial fractions
iii. complete the square, then try a substitution
(c) Radicals
i. trigonometric substitution based on the form of the radical
ii. less obvious rationalizing substitutions (similar to trig sub)
iii. if a quadratic under radical, try completing the square
5. Integration by parts
6. Try substitution or integration by parts again with less obvious choices for u; manipulate the integrand; relate to previous problems; use several methods together.
Z
x
Example 1: ex+e dx
Solution: Notice first that
Z
e
x+ex
Z
dx =
x
ex ee dx
Use the substitution u = ex so that du = ex dx. Thus we have
Z
Z
x
x+ex
e
dx =
ex ee dx
Z
=
eu du
= eu + C
x
= ee + C
Math 262
Page 2
Z
Example 2:
e
√
3
x
dx
√
Solution: Use the rationalizing substitution t = 3 x so that t3 = x and 3t2 dt = dx.
Then we have
Z
Z √
3x
et (3t2 ) dt
e dx =
Z
=
3t2 et dt
Now use integration by parts with
u = 3t2
dv = eu du
so that
v = et
du = 6t dt
Then we have
Z
2 t
2 t
Z
3t e dt = 3t e −
6tet dt
and use integration by parts again with
dv = eu du
u = 6t
so that
v = et
du = 6 dt
Then we have
2 t
3t e −
Z
t
2 t
t
6te dt = 3t e − (6te −
Z
6et dt)
= 3t2 et − 6tet + 6et + C
= (3t2 − 6t + 6)et + C
√
√
√
3
3
= 3e x ( x2 − 2 3 x + 2) + C
Math 262
Page 3
Z
Example 3:
3x2 − 2
dx
x3 − 2x − 8
Solution: Let u = x3 − 2x − 8 so that du = 3x2 − 2 dx. Thus we have
Z
Z
3x2 − 2
1
dx =
du
3
x − 2x − 8
u
= ln |u| + C
= ln |x3 − 2x − 8| + C
Z
Example 4:
3x2 − 2
dx
x2 − 2x − 8
Solution: This is an improper fraction so we divide.
x2 − 2x − 8
So we have that
Notice that
| 3x2
3x2
3
+ 0x − 2
− 6x − 24
6x + 22
3x2 − 2
6x + 22
=3+ 2
2
x − 2x − 8
x − 2x − 8
6x + 22
6x + 22
=
− 2x − 8
(x + 2)(x − 4)
x2
so the partial fraction decomposition will have form
A
B
+
x+2 x−4
A
B
6x + 22
=
+
(x + 2)(x − 4)
x+2 x−4
A(x − 4) + B(x + 2)
=
(x + 2)(x − 4)
6x + 22 = A(x − 4) + B(x + 2)
6x + 22 = Ax − 4A + Bx + 2B
6x + 22 = (A + B)x + (−4A + 2B)
So we know that
A+B =6
and
− 4A + 2B = 22
Math 262
Page 4
which implies that A = − 53 and B =
23
.
3
Thus
23
− 53
6x + 22
=
+ 3
x2 − 2x − 8
x+2 x−4
So we have
Z
3x2 − 2
dx =
x2 − 2x − 8
Z 5
1
23
1
3−
+
dx
3 x+2
3 x−3
23
5
ln |x − 4| + C
= 3x − ln |x + 2| +
3
3
Z
π/3
Example 5:
π/4
ln(tan x)
dx
sin x cos x
2
x
Solution: Let u = ln(tan x) so that du = sec
dx = sin x1cos x dx. Our limits of integratan x
tion become
π
π ln(tan x)
−−−−−→ ln tan
= ln 1 = 0
4
4
√
π
π ln(tan x)
−−−−−→ ln tan
= ln 3
3
3
Thus we have
Z
π/3
π/4
ln(tan x)
dx =
sin x cos x
=
=
=
=
Z
ln
√
3
u du
0
ln √3
1 2 u
2 0
i
1h √ 2
(ln 3) − 0
2
2
1 1
ln 3
2 2
1
(ln 3)2
8
Math 262
Page 5
Z √
Example 6:
1 + ln x
dx
x ln x
Solution: Let u =
Thus we have
√
1 + ln x so that u2 = 1 + ln x, ln x = u2 − 1 and 2u du =
Z √
Z
u
(2u) du
−1
Z
2u2
=
du
u2 − 1
1 + ln x
dx =
x ln x
u2
This is an improper fraction so we divide.
2
2
u − 1 | 2u
2u2
So we have that
+
0u
2
+ 0
− 2
2
2
2u2
=
2
+
u2 − 1
u2 − 1
Notice that
u2
2
2
=
−1
(u + 1)(u − 1)
so the partial fraction decomposition will have form
B
A
+
u+1 u−1
2
A
B
=
+
(u + 1)(u − 1)
u+1 u−1
A(u − 1) + B(u + 1)
=
(u + 1)(u − 1)
2 = A(u − 1) + B(u + 1)
2 = Au − A + Bu + B
2 = (A + B)u + (−A + B)
So we know that
A+B =0
and
−A+B =2
which implies that A = −1 and B = 1. Thus
u2
2
−1
1
=
+
−1
u+1 u−1
1
x
dx.
Math 262
Page 6
So we have
Z
2u2
du =
u2 − 1
=
=
=
=
Z 2
2+ 2
du
u −1
Z 1
1
2−
+
du
u+1 u−1
2u − ln |u + 1| + ln |u − 1| + C
u − 1
+C
2u + ln u + 1
√
√
1 + ln x − 1 +C
2 1 + ln x + ln √
1 + ln x + 1