Integral Calculus

1. Integration of
Trigonometric Functions
RAYMUND T. DE LA CRUZ
MAEd-mathematics

2. Integration by Algebraic Substitution>>>
By noting the formulas for differentiating the six trigonometric functions, and the antiderivatives that are found
using them, we have the following integration formulas:
sin 𝑢 𝑑𝑢 = − cos 𝑢 + 𝐶
cos 𝑢 𝑑𝑢 = sin 𝑢 + 𝐶
sec2 𝑢 𝑑𝑢 = tan 𝑢 + 𝐶
csc2 𝑢 𝑑𝑢 = − cot 𝑢 + 𝐶
sec 𝑢 tan 𝑢 𝑑𝑢 = sec 𝑢 + 𝐶
csc 𝑢 cot 𝑢 𝑑𝑢 = − csc 𝑢 + 𝐶

3. Integration by Algebraic Substitution>>>
Example1
Integrate: 𝑥 𝑠𝑒𝑐2
𝑥2
𝑑𝑥.
With 𝑢 = 𝑥2
, 𝑑𝑢 = 2𝑥 𝑑𝑥, we have
𝑥 𝑠𝑒𝑐2
𝑥2
𝑑𝑥 =
1
2
(𝑠𝑒𝑐2
𝑥2
(2𝑥 𝑑𝑥)
=
1
2
tan 𝑥2
+ 𝐶
Example2
Integrate:
tan 2𝑥
cos 2𝑥
𝑑𝑥.
By using the basic identity sec 𝜃 =
1
cos 𝜃
, we can transform this integral into the form
sec2𝑥 tan2𝑥 𝑑𝑥. In this form, 𝑢 = 2𝑥, 𝑑𝑢 = 2 𝑑𝑥. Therefore,
tan 2𝑥
cos 2𝑥
𝑑𝑥 = sec2𝑥 tan2𝑥 𝑑𝑥 =
1
2
sec2𝑥 tan2𝑥 2 𝑑𝑥
=
1
2
sec2𝑥 + 𝐶

4. Integration by Algebraic Substitution>>>
We also have the following integration formulas:
tan 𝑢 𝑑𝑢 = − 𝑙𝑛 cos 𝑢 + 𝐶
cot 𝑢 𝑑𝑢 = ln sin 𝑢 + 𝐶
sec 𝑢 𝑑𝑢 = ln sec 𝑢 + tan 𝑢 + 𝐶
csc 𝑢 𝑑𝑢 = ln csc 𝑢 − cot 𝑢 + 𝐶
Example3
Integrate: tan 4𝜃 𝑑𝜃.
Noting that 𝑢 = 4𝜃, 𝑑𝑢 = 4𝑑𝜃, we have
tan 4𝜃 𝑑𝜃 =
1
4
tan 4𝜃(4𝑑𝜃)
= −
1
4
ln cos 4𝜃 + 𝐶

5. Integration by Algebraic Substitution>>>
By use of trigonometric relations, it is possible to transform many integrals involving powers of the
trigonometric functions into integrable form. We now show the relationships that are useful for
these integrals.
𝑐𝑜𝑠2
𝑥 + 𝑠𝑖𝑛2
𝑥 = 1
1 + 𝑡𝑎𝑛2
𝑥 = 𝑠𝑒𝑐2
𝑥
1 + 𝑐𝑜𝑡2
𝑥 = 𝑐𝑠𝑐2
𝑥
2𝑐𝑜𝑠2
𝑥 = 1 + cos 2𝑥
2𝑠𝑖𝑛2
𝑥 = 1 − cos 2𝑥
Example1
Integration with odd power of sin u
Integrate: 𝑠𝑖𝑛3
𝑥 𝑐𝑜𝑠2
𝑥 𝑑𝑥.
Because 𝑠𝑖𝑛3
𝑥 = 𝑠𝑖𝑛2
𝑥 sin 𝑥 = 1 − 𝑐𝑜𝑠2
𝑥 sin 𝑥, we can write this integral with powers of
cos x along with – sin x, which is the necessary 𝑑𝑢 for this integral. Therefore,
𝑠𝑖𝑛3
𝑥 𝑐𝑜𝑠2
𝑥 𝑑𝑥 = (1 − 𝑐𝑜𝑠2
𝑥)(sin 𝑥) 𝑐𝑜𝑠2
𝑥 𝑑𝑥
= (𝑐𝑜𝑠2
𝑥 − 𝑐𝑜𝑠4
𝑥)(sin 𝑥 𝑑𝑥)
=− 𝑐𝑜𝑠2
𝑥(− sin 𝑥 𝑑𝑥) + 𝑐𝑜𝑠4
𝑥(− sin 𝑥 𝑑𝑥)
= −
1
3
𝑐𝑜𝑠3
𝑥 +
1
5
𝑐𝑜𝑠5
𝑥 + 𝐶

6. Integration by Algebraic Substitution>>>
Example2
Integration with odd power of cos u
Integrate: 𝑐𝑜𝑠5
2𝑥 𝑑𝑥.
Because 𝑐𝑜𝑠5
2𝑥 = 𝑐𝑜𝑠4
2𝑥 cos 2𝑥 = (1 − 𝑠𝑖𝑛2
2𝑥)2
cos 2𝑥, it is possible to write this integral
with powers of sin 2x along with cos 2x dx. Thus, with the introduction of a factor of 2, (cos 2x)(2dx)
is the necessary 𝑑𝑢 for this integral. Thus,
𝑐𝑜𝑠5
2𝑥 𝑑𝑥 = (1 − 𝑠𝑖𝑛2
2𝑥)2
cos 2𝑥 𝑑𝑥
= 1 − 2𝑠𝑖𝑛2
2𝑥 + 𝑠𝑖𝑛4
2𝑥 cos 2𝑥 𝑑𝑥
= cos 2𝑥 𝑑𝑥 − 2𝑠𝑖𝑛2
2𝑥 cos 2𝑥 𝑑𝑥 + 𝑠𝑖𝑛4
2𝑥 cos 2𝑥 𝑑𝑥
=
1
2
cos 2𝑥 2𝑑𝑥 − 𝑠𝑖𝑛2
2𝑥 (2cos 2𝑥 𝑑𝑥) +
1
2
𝑠𝑖𝑛4
2𝑥 (2cos 2𝑥 𝑑𝑥)
=
1
2
𝑠𝑖𝑛 2𝑥 −
1
3
𝑠𝑖𝑛3
2𝑥 +
1
10
𝑠𝑖𝑛5
2𝑥 + 𝐶

7. Integration by Algebraic Substitution>>>
Example4
Integration with power of sec u
Integrate: 𝑠𝑒𝑐3
𝑡 tan 𝑡 𝑑𝑡.
By writing 𝑠𝑒𝑐3
𝑡 tan 𝑡 as 𝑠𝑒𝑐2
𝑡 (sec 𝑡 tan 𝑡), we can use the sec 𝑡 tan 𝑡 𝑑𝑡 as the 𝑑𝑢 of the
integral. Thus,
𝑠𝑒𝑐3
𝑡 tan 𝑡 𝑑𝑡 = 𝑠𝑒𝑐2
𝑡 (sec 𝑡 tan 𝑡 𝑑𝑡)
=
1
3
𝑠𝑒𝑐3
𝑡 + 𝐶
Example3
Integration with even power of sin u
Integrate: 𝑠𝑖𝑛2
2𝑥 𝑑𝑥.
Since 𝑠𝑖𝑛2
2𝑥 =
1
2
(1 − cos 4𝑥), this integral can be transformed into a form that can be
integrated. Therefore we write
𝑠𝑖𝑛2
2𝑥 𝑑𝑥 =
1
2
(1 − cos 4𝑥) 𝑑𝑥
=
1
2
𝑑𝑥 −
1
8
cos 4𝑥(4𝑑𝑥)
=
𝑥
2
−
1
8
sin 4𝑥 + 𝐶

8. Integration by Algebraic Substitution>>>
Example5
Integration with power of tan u
Integrate: 𝑡𝑎𝑛5
𝑥 𝑑𝑥.
Because 𝑡𝑎𝑛5
𝑥 = 𝑡𝑎𝑛3
𝑥 𝑡𝑎𝑛2
𝑥 = 𝑡𝑎𝑛3
𝑠𝑒𝑐2
− 1 , we can write this integral with powers of
tan 𝑥 along with 𝑠𝑒𝑐2
𝑥 𝑑𝑥. Thus, 𝑠𝑒𝑐2
𝑥 𝑑𝑥 becomes the necessary 𝑑𝑢 of the integral. It is necessary
to replace 𝑡𝑎𝑛2
𝑥 with 𝑠𝑒𝑐2
𝑥 − 1 twice during the integration. Therefore,
𝑡𝑎𝑛5
𝑥 𝑑𝑥 = 𝑡𝑎𝑛3
𝑠𝑒𝑐2
− 1 𝑑𝑥
= 𝑡𝑎𝑛3
𝑥 (𝑠𝑒𝑐2
𝑥 𝑑𝑥) − 𝑡𝑎𝑛3
𝑥 𝑑𝑥
=
1
4
𝑡𝑎𝑛4
𝑥 − tan 𝑥 𝑠𝑒𝑐2
𝑥 − 1 𝑑𝑥
=
1
4
𝑡𝑎𝑛4
𝑥 − tan 𝑥(𝑠𝑒𝑐2
𝑥 𝑑𝑥) + tan 𝑥 𝑑𝑥
=
1
4
𝑡𝑎𝑛4
𝑥 −
1
2
𝑡𝑎𝑛2
𝑥 − ln cos 𝑥 + 𝐶
P r a c t ic e Ex e r c is e
Integrate:
1. sin5𝑥 𝑑𝑥 4. 𝑠𝑖𝑛3
𝑥 𝑑𝑥
2. 6𝑐𝑠𝑐2
3𝑥 𝑑𝑥 5. 𝑠𝑒𝑐4
𝑥 𝑑𝑥
3. 4𝑥 cot 𝑥2
𝑑𝑥

9. Integration by Trigonometric Substitution>>>
Substitutions based on trigonometric relations are particularly useful for integrating expressions
involving radicals.
Example1 For 𝑎2 − 𝑥2, let 𝑥 = 𝑎 sin 𝜃
Integrate:
𝑑𝑥
𝑥2 1−𝑥2
.
If we let 𝑥 = sin 𝜃, then 1 − 𝑠𝑖𝑛2 𝜃 = cos 𝜃, and the integral can be transformed into a
trigonometric integral. Carefully replacing all factors of the integral with expressions in terms of 𝜃,
we have 𝑥 = sin 𝜃, 1 − 𝑥2 = cos 𝜃, and 𝑑𝑥 = cos 𝜃 𝑑𝜃. Therefore,
𝑑𝑥
𝑥2 1−𝑥2
=
cos 𝜃 𝑑𝜃
𝑠𝑖𝑛 2 𝜃 1−𝑠𝑖𝑛 2 𝜃
=
𝑐𝑜𝑠𝜃 𝑑𝜃
𝑠𝑖𝑛 2 𝜃 cos 𝜃
= 𝑐𝑠𝑐2
𝜃 𝑑𝜃
= − cot 𝜃 + 𝐶
We have now performed the integration, but the answer we now have is in terms of 𝜃, and we must
express the result in terms of x. Making a triangle with an angle 𝜃 such that sin 𝜃 =
𝑥
1
, we may
express any of the trigonometric functions in terms of x. Thus,
cot 𝜃 =
1−𝑥2
𝑥
Therefore, the result of the integration becomes
𝑑𝑥
𝑥2 1−𝑥2
= − cot 𝜃 + 𝐶 = −
1−𝑥2
𝑥
+ 𝐶

10. Integration by Trigonometric Substitution>>>
Example2 For 𝑎2 + 𝑥2, let 𝑥 = 𝑎 tan 𝜃
Integrate:
𝑑𝑥
𝑥2+4
.
If we let 𝑥 = 2 tan 𝜃, the radical in this integral becomes
𝑥2 + 4 = 4 𝑡𝑎𝑛2 𝜃 + 4 = 2 𝑡𝑎𝑛2 𝜃 + 1 = 2 𝑠𝑒𝑐2 𝜃 = 2 sec 𝜃
Therefore, with 𝑥 = 2 tan 𝜃 and 𝑑𝑥 = 2 𝑠𝑒𝑐2
𝜃 𝑑𝜃, we have
𝑑𝑥
𝑥2+4
=
2 𝑠𝑒𝑐 2 𝜃 𝑑𝜃
4 𝑡𝑎𝑛 2 𝜃+4
=
2 𝑠𝑒𝑐 2 𝜃 𝑑𝜃
2 sec 𝜃
= sec 𝜃 𝑑𝜃 = ln sec 𝜃 + tan 𝜃 + 𝐶
= ln
𝑥2+4
2
+
𝑥
2
+ 𝐶 = ln
𝑥2+4+𝑥
2
+ 𝐶
This answer is acceptable, but by using the properties of logarithms, we have
ln
𝑥2+4+𝑥
2
+ 𝐶 = ln 𝑥2 + 4 + 𝑥 + (𝐶 − ln 2)
= ln 𝑥2 + 4 + 𝑥 + 𝐶′

11. Integration by Trigonometric Substitution>>>
Example3 For 𝑥2 − 𝑎2, let 𝑥 = 𝑎 sec 𝜃
Integrate:
2 𝑑𝑥
𝑥 𝑥2−9
.
If we let 𝑥 = 3 sec 𝜃, the radical in this integral becomes
𝑥2 − 9 = 9 𝑠𝑒𝑐2 𝜃 − 9 = 3 𝑠𝑒𝑐2 𝜃 − 1 = 3 𝑡𝑎𝑛2 𝜃 = 3 tan 𝜃
Therefore, with 𝑥 = 3 sec 𝜃 tan 𝜃 𝑑𝜃, we have
2 𝑑𝑥
𝑥 𝑥2−9
= 2
3 sec 𝜃 tan 𝜃 𝑑𝜃
3 sec 𝜃 9𝑠𝑒𝑐 2 𝜃−9
= 2
tan 𝜃 𝑑𝜃
3 tan 𝜃
=
2
3
𝑑𝜃 =
2
3
𝜃 + 𝐶 =
2
3
𝑠𝑒𝑐−1 𝑥
3
+ 𝐶
P r a c t ic e Ex e r c is e
Integrate:
1.
1−𝑥2
𝑥2
𝑑𝑥 4.
6𝑑𝑧
𝑧2 𝑧2+9
2.
2𝑑𝑥
𝑥2−36
5.
6𝑑𝑥
𝑥 4−𝑥2
3.
𝑥2−25
𝑥
𝑑𝑥