1. Chapter 8
8.2 Integration By Parts
Let G(x) be any antiderivative of g(x). In this case G’(x)=g(x), then by the product rule,
Which implies
Or, equivalently, as
(1)
The application of this formula is called integration by parts.

2. In practice, we usually rewrite (1) by letting
This yields the following alternative form for (1)

3. Example: Use integration by parts to evaluate
Solution: Let

4. Guidelines for Integration by Parts
The main goal in integration by parts is to choose u and dv to obtain a new integral
That is easier to evaluate than the original.
A strategy that often works is to choose u and dv so that u becomes “simpler” when
Differentiated, while leaving a dv that can be readily integrated to obtain v.
There is another useful strategy for choosing u and dv that can be applied when the
Integrand is a product of two functions from different categories in the list.
Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential
In this case, you will often be successful if you take u to be the function whose
Category occurs earlier in the list and take dv to the rest of the integrand (LIATE).
This method does not work all the time, but it works often enough to b e useful.

5. Example. Evaluate
Solution: According to LIATE, we should let

6. Example: Evaluate
Solution: Let

7. Repeated Integration by Parts
Example: Evaluate
Solution: Let
Apply integration by parts to

8. Finally,

9. Example: Evaluate
Solution: Let
(a)

10. Together with (a), we have
Solve for the unknown integral, we have

11. Integration by Parts for Definite Integrals
For definite integrals, the formula corresponding to is

12. Example: Evaluate
Solution: Let

13. 8.3 Trigonometric Integrals
Integrating powers of sine and cosine
By applying the integration by parts , we have two reduction formulas

14. In particular, ……

15. Integrating Products of Sines and Cosines

16. Example: Evaluate
Solution: since n=5 is odd,

17. Integrating Powers of Tangent and Secant
There are similar reduction formulas to integrate powers of tangent and secant.

18. In particular, ……

19. Integrating Products of Tangents and Secents (optional)

20. Section 8.4 Trigonometric Substitutions
We are concerned with integrals that contain expressions of the form
In which a is a positive constant. The idea is to make a substitution for x that will eliminate the radical.
For example, to eliminate the radical in , we can make the substitution
x = a sinx, - /2    /2
Which yields

21. Example Example 1. Evaluate
Solution. To eliminate the radical we make the substitution
x= 2sin, dx= 2cosd
This yields

22. Example Cont. From the left figure, we obtain
Substituting this in the previous answer yields

23. Example There are two methods to evaluate the definite integral.
Make the substitution in the indefinite integral and then evaluate the definite integral using the x-limits of integration.
Make the substitution in the definite integral and convert the x-limits to the corresponding -limits.
Example. Evaluate
Solution: Since the substitution can be expressed as =sin-1(x/2), the -limits of the integration are
x=1: = sin-1(1/2)=/6
x= : = sin-1( /2)=/4
Thus

24. Method of Trigonometric Substitution

25. Example dx= 5sec tan d. Thus,
Example. Evaluate , assuming that x5.
Solution. Let x=5 sec, 0  </2
dx= 5sec tan d.
Thus,
Note that

26. 8.5 Integrating Rational Functions by Partial Fractionas
Suppose that P(x) / Q(x) is a proper rational function, by which we mean that the
Degreee of the numerator is less than the degree of the denominator. There is a
Theorem in advanced algebra which states that every proper rational function can
Be expressed as a sum
Where F1(x), F2(x), …, Fn(x) are rational functions of the form
In which the denominators are factors of Q(x). The sum is called the partial fraction
Decomposition of P(x)/Q(x), and the terms are called partial fraction.

27. Finding the Form of a Partial Fraction Decomposition
The first step in finding the form of the partial fraction decomposition of a proper
Rational function P(x)/Q(x) is to factor Q(x) completely into linear and irreducible
Quadratic factors, and then collect all repeated factors so that Q(x) is expressed as
A product of distinct factors of the form
From these factors we can determine the form of the partial fraction decomposition
Using two rules that we will now discuss.

28. Linear Factors
If all of the factors of Q(x) are linear, then the partial fraction decomposition of
P(x)/Q(x) can be determined by using the following rule:

29. Example: Evaluate
Solution:
By the linear factor rule, the decomposition has the form
Where A and B are constants to be determined. Multiplying this expression
Through by (x-1)(x+2) yields
1 = A ( x + 2 ) + B ( x – 1 )
Let x=1, we have 1= 3A which implies A = 1/3; and let x = -2, we have 1 = -3B,
Which means B = -1/3. Thus,

30. The integration can now be completed as follows:

31. Example: Evaluate
Solution:
By the linear factor rule,
Multiplying by yields
(1)
(2)
Setting x=0 in (1) yields B = -2, and setting x = 2 in (1) yields C = 2. Then
By equating the coefficient of x2 on the two sides to obtain
A + C = 0 or A = -C = -2

32. Substituting A = -2, B = -2, and C = 2 yields the partial fraction decomposition
Thus

33. Quadratic Factors
If some of the factors of Q (x) are irreducible quadratics, then the contribution of
Those facctors to the partial fraction decomposition of P(x)/Q(x) can be determined
From the following rule: