1. 5
INTEGRALS

2. INTEGRALS
Indefinite Integrals

3. Thus, ∫ f(x) dx = F(x) means F’(x) = f(x)
INDEFINITE INTEGRAL
The notation ∫ f(x) dx is traditionally used for an antiderivative of f and is called an indefinite integral.
Thus, ∫ f(x) dx = F(x) means F’(x) = f(x)

4. For example, we can write
INDEFINITE INTEGRALS
For example, we can write
Thus, we can regard an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C).

5. INDEFINITE INTEGRALS
Any formula can be verified by differentiating the function on the right side and obtaining the integrand.
For instance,

6. TABLE OF INDEFINITE INTEGRALS

7. INDEFINITE INTEGRALS
Thus, we write with the understanding that it is valid on the interval (0, ∞) or on the interval (-∞, 0).

8. INDEFINITE INTEGRALS
This is true despite the fact that the general antiderivative of the function f(x) = 1/x2, x ≠ 0, is:

9. Find the general indefinite integral ∫ (10x4 – 2 sec2x) dx
INDEFINITE INTEGRALS
Example 1
Find the general indefinite integral ∫ (10x4 – 2 sec2x) dx
Using our convention and Table 1, we have: ∫(10x4 – 2 sec2x) dx = 10 ∫ x4 dx – 2 ∫ sec2x dx = 10(x5/5) – 2 tan x + C = 2x5 – 2 tan x + C
You should check this answer by differentiating it.

10. Evaluate INDEFINITE INTEGRALS Example 2
This indefinite integral isn’t immediately apparent in Table 1.
So, we use trigonometric identities to rewrite the function before integrating:

11. The Substitution Rule INTEGRALS In this section, we will learn:
To substitute a new variable in place of an existing
expression in a function, making integration easier.

12. INTRODUCTION
Equation 1
Our antidifferentiation formulas don’t tell us how to evaluate integrals such as

13. INTRODUCTION
To find this integral, we use the problem-solving strategy of introducing something extra.
The ‘something extra’ is a new variable.
We change from the variable x to a new variable u.

14. INTRODUCTION
Suppose we let u = 1 + x2 be the quantity under the root sign in the integral below,
Then, the differential of u is du = 2x dx.

15. INTRODUCTION
Notice that, if the dx in the notation for an integral were to be interpreted as a differential, then the differential 2x dx would occur in

16. So, formally, without justifying our calculation, we could write:
INTRODUCTION
Equation 2
So, formally, without justifying our calculation, we could write:

17. INTRODUCTION
However, now we can check that we have the correct answer by using the Chain Rule to differentiate

18. INTRODUCTION
In general, this method works whenever we have an integral that we can write in the form ∫ f(g(x))g’(x) dx

19. ∫ F’(g(x))g’(x) dx = F(g(x)) + C
INTRODUCTION
Equation 3
Observe that, if F’ = f, then
∫ F’(g(x))g’(x) dx = F(g(x)) + C
because, by the Chain Rule,

20. INTRODUCTION
If we make the ‘change of variable’ or ‘substitution’ u = g(x), we have:

21. Writing F’ = f, we get: ∫ f(g(x))g’(x) dx = ∫ f(u) du
INTRODUCTION
Writing F’ = f, we get: ∫ f(g(x))g’(x) dx = ∫ f(u) du
Thus, we have proved the following rule.

22. SUBSTITUTION RULE
Equation 4
If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫ f(g(x))g’(x) dx = ∫ f(u) du

23. Notice also that, if u = g(x), then du = g’(x) dx.
SUBSTITUTION RULE
Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation.
Notice also that, if u = g(x), then du = g’(x) dx.
So, a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.

24. Thus, the Substitution Rule says:
It is permissible to operate with dx and du after integral signs as if they were differentials.

25. Find ∫ x3 cos(x4 + 2) dx We make the substitution u = x4 + 2.
SUBSTITUTION RULE
Example 1
Find ∫ x3 cos(x4 + 2) dx
We make the substitution u = x4 + 2.
This is because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral.

26. Thus, using x3 dx = du/4 and the Substitution Rule, we have:
Example 1
Thus, using x3 dx = du/4 and the Substitution Rule, we have:
Notice that, at the final stage, we had to return to the original variable x.

27. SUBSTITUTION RULE
The idea behind the Substitution Rule is to replace a relatively complicated integral by a simpler integral.
This is accomplished by changing from the original variable x to a new variable u that is a function of x.
Thus, in Example 1, we replaced the integral ∫ x3cos(x4 + 2) dx by the simpler integral ¼ ∫ cos u du.

28. SUBSTITUTION RULE
The main challenge in using the rule is to think of an appropriate substitution.
You should try to choose u to be some function in the integrand whose differential also occurs—except for a constant factor.
This was the case in Example 1.

29. SUBSTITUTION RULE
If that is not possible, try choosing u to be some complicated part of the integrand—perhaps the inner function in a composite function.

30. Finding the right substitution is a bit of an art.
SUBSTITUTION RULE
Finding the right substitution is a bit of an art.
It’s not unusual to guess wrong.
If your first guess doesn’t work, try another substitution.

31. Evaluate Let u = 2x + 1. Then, du = 2 dx. So, dx = du/2.
SUBSTITUTION RULE
E. g. 2—Solution 1
Evaluate
Let u = 2x + 1.
Then, du = 2 dx.
So, dx = du/2.

32. SUBSTITUTION RULE
E. g. 2—Solution 1
Thus, the rule gives:

33. Find SUBSTITUTION RULE Example 3 Let u = 1 – 4x2. Then, du = -8x dx.
So, x dx = -1/8 du and

34. Calculate ∫ cos 5x dx If we let u = 5x, then du = 5 dx.
SUBSTITUTION RULE
Example 4
Calculate ∫ cos 5x dx
If we let u = 5x, then du = 5 dx.
So, dx = 1/5 du.
Therefore,