The document summarizes additional integration techniques presented in Chapter 4, beyond those introduced in previous chapters. These techniques include integration by parts, integration of trigonometric expressions, integration by trigonometric substitution, and integration of functions involving quadratics, rationals, or requiring rationalizing substitutions. Examples are provided to demonstrate each technique. The chapter also lists helpful indefinite integral formulas for reference. Exercises are provided to allow readers to practice applying the new integration techniques.

1. 4. Techniques of Integration
(MAT060 - Calculus with Analytic Geometry I)
RAYLEE J. GASPARIN, PhD
Mathematics Department
Mindanao State University Main Campus
Marawi City
gasparin.raylee@msumain.edu.ph
December 8, 2021

2. Chapter 4 . Techniques of Integration
We already learned how to integrate simple algebraic functions,
trigonometric functions, exponential and logarithmic functions and inverse
trigonometric functions by using the method of Integration by
Substitution or by simply using the basic formulas for integration.
In this chapter, additional Techniques of Integration will be
presented. Such techniques are:

3. 1. Integration by parts,
2. Integration of trigonometric expressions,
3. Integration by trigonometric substitution,
4. Integration of integral involving quadratic function,
5. Integration of rational function, and
6. Integration by rationalizing substitution.
Indenite integration formulas that you learned in previous chapter and that
are used frequently are listed below for reference.

4. Some Helpful Formulas
1.
Z
du = u + C
2.
Z
adu = au = C where a is any constant
3.
Z
[f(u) + g(u)] =
Z
f(u)du +
Z
g(u)du
4.
Z
un
du =
un+1
n + 1
+ C , n 6= −1
5.
Z
du
u
= ln |u| + C
6.
Z
eu
du = eu
+ C
7.
Z
au
du =
au
ln a
+ C, where a 0 and a 6= 1

5. Some Helpful Trigonometric Function Formulas
1.
Z
sin u du = − cos u + C
2.
Z
cos u du = sin u + C
3.
Z
sec2
u du = tan u + C
4.
Z
csc2
u du = − cot u + C
5.
Z
sec u tan u du = sec u + C
6.
Z
csc u cot u du = − csc u + C
7.
Z
tan u du = ln | sec u| + C
8.
Z
cot u du = − ln | csc u| + C
9.
Z
sec u du = ln | sec u + tan u| + C
10.
Z
csc u du = − ln | csc u + cot u| + C

6. Integration Yielding Inverse Trigonometric Functions:
1.
Z
du
√
a2 − u2
= arcsin
u
a
+ C
2.
Z
du
a2 + u2
=
1
a
arctan
u
a
+ C
3.
Z
du
u
√
u2 − a2
=
1
a
arcsec
u
a
+ C

7. 4.1 The Integration by Parts
The formula for the dierential of a product is
d(uv) = udv + vdu.
By integrating both sides, we obtain
uv =
Z
udv +
Z
vdu. (1)
From (1) the following formula for integration by parts,
Z
udv = uv −
Z
vdu.

8. Thus, Integration by Parts is the process which involves splitting the
integrand into two parts, u and dv where
R
vdu is hoped to be a simpler
integral.

9. Example 24.1 Find
Z
ln xdx.
Solution: We let u = ln x and dv = dx. Then du =
dx
x
and v = x.
So,
Z
ln xdx = x ln x −
Z
x
dx
x
= x ln x −
Z
dx
= x ln x − x + C.

10. Example 24.2 Evaluate
Z
x3
ex2
dx.
Solution: We write
Z
x3
ex2
dx =
Z
x2
ex2
xdx
and let u = x2
,
dv = xex2
dx. Then du = 2xdx and v =
1
2
ex2
. Thus,
Z
x3
ex2
dx =
Z
x2
ex2
xdx
=
1
2
x2
ex2
−
Z
1
2
ex2
(2xdx)

11. =
x2
ex2
2
−
Z
ex2
(xdx)
=
x2
ex2
2
−
1
2
ex2
+ C
Z
x3
ex2
dx =
ex2
2
x2
− 1
+ C.

12. Example 24.3 Find
Z
ex
cos xdx.
Solution: Let u = ex
and dv = cos xdx. Then du = ex
dx and v = sin x.
So, Z
ex
cos xdx = ex
sin x −
Z
ex
sin xdx.
Observe that
Z
ex
sin xdx cannot be directly or easily integrated. We need
to perform integration by parts again.
Let u2 = ex
and dv2 = sin xdx. Then du2 = ex
dx and v2 = − cos x.
We have,

13. Z
ex
cos xdx = ex
sin x −
Z
ex
sin xdx
= ex
sin x −
−ex
cos x −
Z
−ex
cos xdx
= ex
sin x + ex
cos x −
Z
ex
cos xdx
2
Z
ex
cos xdx = ex
sin x + ex
cos x
Therefore, Z
ex
cos xdx =
1
2
[ex
sin x + ex
cos x] .

14. Example 24.4 Find
Z
x arcsec xdx.
Solution: Let u = arcsec x and dv = xdx. Then du =
dx
x
√
x2 − 1
and
v =
x2
2
. So,
Z
x arcsec xdx =
x2
2
arcsec x −
Z
x2
2
·
dx
x
√
x2 − 1

15. =
x2
2
arcsec x −
1
2
Z
xdx
x
√
x2 − 1
=
x2
2
arcsec x −
1
4
2
√
x2 − 1
+ C
Z
x arcsec xdx =
x2
2
arcsec x −
1
2
√
x2 − 1 + C

16. Exercises: Determine the following integrals:
1.
Z
x sec2
xdx
2.
Z
arctan xdx
3.
Z
x3
e2x
dx
4.
Z
sin(2x) cos(3x) dx
5.
Z
csc3
(2x) dx
6.
Z 2
1
e−x
ln x dx
7.
Z
3w2
arccos(2w) dw