1. Integration Techniques
A. Trigonometric Substitution
B. Rational Functions / Partial
Fractions
C. Rationalizing Substitution
2. Improper Integrals
3. Application of Definite Integrals
Integral
Calculus
Learning module
Aero Department
Integration Techniques- Trigonometric Substitution
I. TITLE:
Integration Techniques- Trigonometric Substitution
Some integrals involving trigonometric functions can be evaluated by using the trigonometric identities. These allow
the integrand to be written in an alternative form which may be more amenable to integration.
II. OBJECTIVES:
At the end of this module, students should be able to:
a. Perform integration using Trigonometric Substitution.
b. Distinguish problems which can be integrated using Trigonometric Substitution.
III. INTRODUCTION:
As discussed, the problems involving inverse trigonometric functions may be solved using the base integrals given in
the previous lectures. As convenient as it may seem, the method presented difficulty and confusion in the entire
process of solving. In line with this, an integration technique known as trigonometric substitution may be used in
solving problems involving terms such as:
IV. PRE-REQUISITE:
To be able to fully understand this module, we must be familiar with the lecture on inverse trigonometric function:
1. ∫
√
( )
2. ∫
3. ∫
√
( )
V. IMPORTANT CONSIDERATIONS:
As for trigonometric substitution, we just need to take note of the following conventions to use similar to the
integration of inverse trigonometric functions:
1. When the integrand involves ,
2. When the integrand involves,
3. When the integrand involves,
In addition, basic trigonometric identities shall be used in the process of trigonometric substitution:
1.
2.
3.
VI. STEPS TO FOLLOW WHEN USING TRIGONOMETRIC SUBSTITUTION:
1.
2.
3.
4.
5.
6.
Determine the term involved in line with the convention to use
Replace the function with the equivalent trigonometric function
Reduce that term by factoring the constant a and reduce the factored term in an equivalent identity
Replace every term left in terms of x or dx with the derivative of
,
, and
.
Integrate as if you are integrating basic trigonometric function
Construct a right triangle conforming to the convention used and get the last side missing in the triangle
using Pythagorean theorem.
7. Replace everything in terms of trigonometric function back to the original variables and constant used using
the right triangle.
VII. EXAMPLE:
1. ∫
SOLUTION:
Step 1: Since it involves
, we will use
Step 2: Replace x with
to get:
∫
Step 3: Factor out a and reduce using equivalent identity:
=∫
∫
=∫
=∫
Step 4: Replace remaining term in terms of x and dx with an equivalent trigonometric function by deriving
initial assumption that
which has a derivative of
=∫
∫
=
=
∫
∫
Step 5: Integrate as if you are integrating basic trigonometric function
∫
=
BUT it is still in terms of theta, we need to replace it in terms of x.
Step 6: Construct a right triangle conforming to the convention used
side is equal to √
; where
Missing
using Pythagorean theorem.
Step 7: Replace everything in terms of trigonometric function back to the original variables and constant
used using the right triangle. Where tan
=
(
√
)
=(
=
√
using the triangle.
√
)
VIII. REFERENCES:





Differential and Integral Calculus 6th Edition, Love & Rainville
https://www.youtube.com/watch?v=3lC5AuCFK4c
https://www.youtube.com/watch?v=GVL4lHX6DgM
https://www.youtube.com/watch?v=7QxAx4Y2RvE&t=20s
https://www.youtube.com/watch?v=yW6Odu0YHL0
 https://www.youtube.com/watch?v=wuRNmNDLanc
 https://www.youtube.com/watch?v=gJdeJ1CoFnU
IX. Practice Problems:
1. ∫
2. ∫
3. ∫
4. ∫
√
√
√
Integration Techniques-Rational Functions / Partial Fractions
I. TITLE:
Integration Techniques-Rational Functions / Partial Fractions
Integration by partial fractions involves terms called rational fractions which are terms that have lower degree in the
numerator than in the denominator. This is the first thing we should check if it seems that the problem would qualify
for partial fraction decomposition.
II. OBJECTIVES:
At the end of this module, students should be able to:
c. Perform integration using Partial Fractions.
d. Distinguish problems which can be integrated using Partial Fractions.
III. IMPORTANT CONSIDERATIONS
Partial fractions is the name given to a technique of integration that may be used to integrate any ratio of
polynomials. A ratio of polynomials is called a rational function. Suppose that N(x) and D(x) are
polynomials. The basic strategy is to write N(x) as a sum of very simple, easy to integrate rational functions,
D(x)
namely
 Polynomials (which are needed only if the degree(1) of N(x) is equal to or strictly bigger than the
degree of D(x)) and
 Rational functions of the particularly simple form

Rational functions of the form
and
. (We’ll see how to integrate them shortly.)
If the degree in the numerator is higher than in the denominator, use polynomial or synthetic division first.
IV. CASES OF PARTIAL FRACTIONS

CASE I: DISTINCT LINEAR FACTORS
- Each of the factors get their respective linear constant
-∫

∫(
)
CASE III: DISTINCT QUADRATIC FACTORS
- Each of the quadratic factors get linear factor (one degree less)
-∫

)
CASE II: LINEAR FACTORS, SOME REPEATED
- Build each of the repeated linear factors and get separate linear constants for each factors
-∫

∫(
∫(
)
CASE IV: QUADRATIC FACTORS, SOME REPEATED
- Build each of the repeated quadratic factors and get separate linear factors for each
-∫
∫(
)
V. STEPS TO FOLLOW WHEN USING PARTIAL FRACTION DECOMPOSITION:
1. Evaluate the rational function and determine to which case it applies.
2. Perform the factor evaluation and get the constants in it. (A,B,C,D,…)
3. After the constant is determined, perform integration with each respective term.
VI. EXAMPLE:
2. ∫
SOLUTION:
Step 1: The problem applies to case I, therefore:
∫(
Let: ∫
)
Step 2: Perform the factor evaluation and get the constants in it. We should get the value of A and B to get
the equivalent function of the problem. We should remove our integral sign and dx during the evaluation of
A and B.
(
)
*Multiply both sides by x and x+1 to get:
Expand the term:
Group the terms with the same variable:
Equate the terms with the same variables:
0x +
* 0x = (A+B)x
*1=A
From the equation above we get the value of A = 1. To get B we should cancel x on both sides to get:
0 = A + B ; where A=1
B = -1
Step 3: Since all the constants are evaluated we could return to our equivalent function where A = 1 and B = 1.
∫
∫(
∫
∫(
)
)
Integrate the equivalent function:
∫(
)
(
)
*Solving A and B using different method
From the previous equation;
We must assume values of x that will make A and/or B multiplied by zero to cancel these variable. In this
case we assume x as equals 0 and -1.
For x=0,
(
For x=-1
)
We can see that our values of A and B are the same.
VII. REFERENCES:

Differential and Integral Calculus 6th Edition, Love & Rainville

https://www.youtube.com/watch?v=BvGKVn-85jM

https://www.youtube.com/watch?v=r5MIraVUVUA

https://www.youtube.com/watch?v=HZTv4zCgEnA

https://www.youtube.com/watch?v=Slfrb8aEj08
VIII. PRACTICE PROBLEMS:
1. ∫
2. ∫
3. ∫
4. ∫
5. ∫
(
)
(
)
Integration Technique – Rationalizing Substitution
I.
TITLE:
Integration Technique – Rationalizing Substitution
Integration by Rationalizing Substitution involves a typical u-substitution provided that the fractional powers of the
respective variable may be reduced into something we can work on and integrate with ease. Therefore, this method
emphasizes its use on elimination of different powers of radicals.
II.
OBJECTIVES:
At the end of this module, students should be able to:
A. Perform integration using Rationalizing Substitution.
B. Distinguish problems which can be integrated using Rationalizing Substitution.
III.
STEPS TO FOLLOW WHEN USING RATIONALIZING SUBSTITUTION:
1. Let u be any powers of the respective variable in order to eliminate is power (which is true for radical
expressions)
2. Derive u and replace everything with respect to the initial variable used in the given problem with respect to
your initial substitution in step 1
IV.
1. ∫
EXAMPLE:
√
√
SOLUTION:
Step 1: Perform U-Substitution to eliminate radical power of x
Let:
√
√
Step 2: Derive u and replace every term with the equivalent substitution performed
∫
√
√
=∫
= ∫
By long division we could further simplify to get an integrand of this form:
∫(
)
Integrating the first three terms by power rule and the last term with arctan we will get:
[
Replace u in terms of x with
]
;
[
to get the final answer of:
√
(
)
]
V.





REFERENCES:
Differential and Integral Calculus 6th Edition, Love & Rainville
https://www.youtube.com/watch?v=Wyh7mSsSQYw&t=508s
https://www.youtube.com/watch?v=jXWbL7FbQWc
https://www.youtube.com/watch?v=D8OncQ9-dHY
https://www.youtube.com/watch?v=BdlI4g0p70w
VI.
PRACTICE PROBLEMS:
1. ∫
2. ∫
3. ∫
√
√
√
Improper Integrals
I.
TITLE
Improper Integrals
Improper integrals are definite integrals by nature except for the fact that they are either not bounded or not finite.
Since definite integrals are evaluated from limit a to b, we get a solution in a particular integral at ease but for the
case of improper integrals one of the limits may be extending up to positive or negative infinity. The limit that
extends to an infinite value will give us a problem in arriving at finite solution that we need to evaluate a particular
problem.
A converging function is defined as a function whose value as a whole trends to a decreasing one where as a
diverging function is a function whose value has a trend to an increasing one, or pulsating to a higher to a lower one
(or vice versa) infinitely.
The general mechanics for improper integrals involves the principle of limits discussed in Calculus I. Evaluating
improper integrals involves the same steps used in evaluating definite integrals except that either one of the limits
are assumed as a finite limit which approaches infinity.
II.
OBJECTIVES:
At the end of this module, students should be able to:
a. Perform integration involving improper integrals.
b. Evaluate particular integral problems and determine if the equation is converging or diverging.
III.
STEPS TO FOLLOW WHEN EVALUATING IMPROPER INTEGRALS:
1.
2.
3.
4.
Let the infinite limit be assumed as a finite one.
Set the limit of the new limit that approaches either positive or negative infinity.
Evaluate the integral as a definite integral.
Evaluate the limits with the same steps involved in Calculus I (which can include Lhopital’s Rule for
indeterminate forms).
5. Determine if it is divergent or convergent.
IV.
EXAMPLE:
1. ∫
SOLUTION:
Step 1: Let infinity be b
∫
=∫
Step 2: Set the limits of b
∫
=
∫
Step 3: Evaluate as a normal definite integral
∫
*
=
+
=
*
+
Step 4: Evaluate the limits, since b approaches infinity
(but
)
=
=1
Step 5: Since the answer is equal to 1 (or any value greater than 0, i.e. finite)
The integral is 1 and convergent.
2. ∫
SOLUTION:
Step 1: Let infinity be b
∫
=∫
Step 2: Set the limits of b
∫
=
∫
Step 3: Evaluate as a normal definite integral
∫
=
[
]
=
Step 4: Evaluate the limits, since b approaches infinity
=
but
and 3* =
=
Step 5: Since the answer is equal to
(i.e. infinite)
The integral is and divergent.
V.
REFERENCES:
 Differential and Integral Calculus 6th Edition, Love & Rainville
 https://www.youtube.com/watch?v=g-M8FHslgdk
VI.
1. ∫
2. ∫
3. ∫
4. ∫
5. ∫
PRACTICE PROBLEMS:
[
] ; but ln 1 = 0
Application of Definite Integrals
I.
TITLE
Application of Definite Integrals
Integration of definite integrals gives values accurate enough as an estimate for the area under each curve. The
application of definite integrals involves getting areas of shapes that do not have the same geometric properties as
the existing shapes that have their own formulas.
The area of a given function results from the integration of the said function with respect to its limits. The process of
integration estimates the value of an area under the particular curve without having to construct rectangles and
getting the sum of each individual strips. Note: No area should have a negative sign.
The area between curves can be determined by subtracting the area under rightmost curve to the area under the
leftmost curve (or the difference between the upper and lower curves). In general, the first step in determining areas
a vertical or horizontal strip must be constructed. In line with this, it is imperative that the strip touches each of the
curves on each ends. The last parameter needed would be the limits that are given by the intersection of the curves
and can be determined by substitution either of the given function to get a limit in the ordinate or abscissa.
II.
OBJECTIVES:
At the end of this module, students should be able to:
a. Perform integration involving application of definite integral to get plane areas and area between curves.
b. Distinguish each problem and evaluate which method to use.
III.
1.
2.
3.
4.
STEPS TO FOLLOW WHEN GETTING AREAS UNDER THE CURVE:
Make a rough sketch to see what is going on the problem
Create a horizontal or vertical strip to be used
Identify the limits that bounds the curve
Integrate the problem to determine the area
IV.
STEPS TO FOLLOW WHEN GETTING AREAS BETWEEN CURVES:
1. Make a rough sketch to see what is going on the problem
2. Create a horizontal or vertical strip to be used
3. Identify the limits or intersection points of the two curves by equating each curve with each other to get
values in the respective axis
4. Formulate the integrand as the difference of the given curves before integrating to get the area
Note: Please familiarize graphs of lines and curves which include conics, exponential functions, logarithmic functions
and polynomials.
V.






REFERENCES:
Differential and Integral Calculus 6th Edition, Love & Rainville
https://www.youtube.com/watch?v=hxcUaUkzi-0
https://www.youtube.com/watch?v=wXMr2YYDMiM
https://www.youtube.com/watch?v=c7wur9Lixb0
https://www.youtube.com/watch?v=DRFyNHdVgUA
https://www.onlinemathlearning.com/area-under-curve.html
VI.
PRACTICE PROBLEMS:
1. Find the area of the region bounded above by y = x2 + 1, bounded below by y = x, and bounded on the
sides by x = 0 and x = 1.
2. Find the area between the two curves y = x2 and y = 2x – x2
Supplemental Problems
1. ∫
√
2. ∫
3. ∫
14. ∫
15. ∫
√
16. ∫
√
4. ∫
17. ∫
5. ∫
18. ∫
6. ∫
19. ∫
7. ∫
8. ∫
√
(
√ )
21. ∫
22. ∫
23. ∫
√
24. ∫
11. ∫
25. ∫
12. ∫
13. ∫
)
20. ∫
√
9. ∫
10. ∫
(
26. ∫
(
)
27. Find the area of the region bounded above by y = x2 + 1, bounded below by y = x, and bounded on the sides
by x = 0 and x = 1.
28. Find the area between the two curves y = x2 and y = 2x – x2
29. Find the are in the second quadrant bounded by the curve: y = x2 + 1
30. Find the area of a circle with its center located at the origin with a radius of 4 units. (Use integration for your
solution)
31. Find the area of an ellipse with its center located at (4,3) with minor axis length of 4 and a major axis length
of 6. Assume the major axis is parallel to the abscissa. (Use integration for your solution)
32. Suppose that we are taking the area of an aircraft vertical stabilizer plotted within the boundaries of the
following lines and points: (Use integration for your solution)
a. y=1
b. y=7
c. x=2
d. x=8
e. m=2 ; @ (5,7) and (2,1)
f. m=3 ; @ (8,7) and (6,1)