Integration Techniques

Many integration formulas can be derived directly from their corresponding derivative formulas, while other integration problems require more work. Some that require more work are substitution and change of variables, integration by parts, trigonometric integrals, and trigonometric substitutions.

Most of the following basic formulas directly follow the differentiation rules.

1. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 

EXAMPLES:

Example 1: Evaluate 

Using formula (4) from the preceding list, you find that .

Example 2: Evaluate .

Because  using formula (4) from the preceding list yields 

Example 3: Evaluate 

Applying formulas (1), (2), (3), and (4), you find that 

Example 4: Evaluate 

Using formula (13), you find that 

Example 5: Evaluate 

Using formula (19) with a = 5, you find that 

Substitution and change of variables

One of the integration techniques that is useful in evaluating indefinite integrals that do not seem to fit the basic formulas is substitution and change of variables.This technique is often compared to the chain rule for differentiation because they both apply to composite functions. In this method, the inside function of the composition is usually replaced by a single variable (often u). Note that the derivative or a constant multiple of the derivative of the inside function must be a factor of the integrand.

The purpose in using the substitution technique is to rewrite the integration problem in terms of the new variable so that one or more of the basic integration formulas can then be applied. Although this approach may seem like more work initially, it will eventually make the indefinite integral much easier to evaluate.

Note that for the final answer to make sense, it must be written in terms of the original variable of integration.

Example 6: Evaluate 

Because the inside function of the composition is x³ + 1, substitute with 

Example 7: 

Because the inside function of the composition is 5 x, substitute with 

Example 8: Evaluate 

Because the inside function of the composition is 9 – x², substitute with 

Integration by parts

Another integration technique to consider in evaluating indefinite integrals that do not fit the basic formulas is integration by parts. You may consider this method when the integrand is a single transcendental function or a product of an algebraic function and a transcendental function. The basic formula for integration by parts is

where u and v are differential functions of the variable of integration.

A general rule of thumb to follow is to first choose dv as the most complicated part of the integrand that can be easily integrated to find v. The u function will be the remaining part of the integrand that will be differentiated to find du. The goal of this technique is to find an integral, ∫ v du, which is easier to evaluate than the original integral.

Example 9: Evaluate ∫ x sec² x dx. 

Example 10: Evaluate ∫ x⁴ In x dx. 

Example 11: Evaluate ∫ arctan x dx. 

Trigonometric integrals

Integrals involving powers of the trigonometric functions must often be manipulated to get them into a form in which the basic integration formulas can be applied. It is extremely important for you to be familiar with the basic trigonometric identities, because you often used these to rewrite the integrand in a more workable form. As in integration by parts, the goal is to find an integral that is easier to evaluate than the original integral.

Example 12: Evaluate ∫ cos³ x sin⁴ x dx 

Example 13: Evaluate ∫ sec⁶ x dx 

Example 14: Evaluate ∫ sin⁴ x dx 

Trigonometric substitutions

If an integrand contains a radical expression of the form a specific trigonometric substitution may be helpful in evaluating the indefinite integral. Some general rules to follow are

1. If the integrand contains  
   
   
2. If the integrand contains  
   
   
3. If the integrand contains  
   
   

Right triangles may be used in each of the three preceding cases to determine the expression for any of the six trigonometric functions that appear in the evaluation of the indefinite integral.

Example 15: Evaluate 

Because the radical has the form  

Figure 1 Diagram for Example 15.

Example 16: Evaluate 

Because the radical has the form  

Figure 2 Diagram for Example 16.

AREA UNDER A CURVE

Area Integral Approximations

The area under any continuous curve can be approximated by drawing a number of rectangles. The integral is the limit for an infinite number of rectangles.

Area Integral Example

Integrals are useful for finding the area under curves which can be approximated by geometrical methods.

This is a polynomial type integral which can be found as the sum of the integrals of its parts.

The area under a curve between two points can be found by doing a definite integral between the two points.

To find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b.

Areas under the x-axis will come out negative and areas above the x-axis will be positive. This means that you have to be careful when finding an area which is partly above and partly below the x-axis.

You may also be asked to find the area between the curve and the y-axis. To do this, integrate with respect to y.

Example

Find the area bounded by the lines y = 0, y = 1 and y = x².

Area under a curve related to different axes

Example #1

Find the area 'A' enclosed by the x-axis, x=2, x=4 and the graph of y=x³/10.

Positive and negative area

note: This expression calculates the absolute area between the curve the vertical lines at 'a' and 'b' and the x-axis. It takes no account of sign. If sign were an issue then the two integrals on the first line would be added and not subtracted.

Unless told differently, assume that the absolute area is required.

Example #1

Find the area 'A' enclosed by the x-axis, x=1, x=8 and the graph of y=2sin[(x+3)/2].

The curve crosses the x-axis at y=0.

Therefore 2sin[(x+3)/2]=0

Sine is zero when the angle is 0,180 or 360 deg.

(zero, pi and 2 pi)

Area bounded by two curves

Example #1

To 3 d.p. calculate the area 'A' included between the curves y=x²/2 and y=(0.75)x

first find the x value where the curves cross

The area 'A' is the difference between the area under the straight line and the area under the parabola, from x=0 to x=3.5 .

1. Volume of a wine cask

A wine cask has a radius at the top of 30 cm and a radius at the middle of 40 cm. The height of the cask is 1 m. What is the volume of the cask (in L), assuming that the shape of the sides is parabolic?

We will lay the cask on its side to make the algebra easier:

We need to find the equation of a parabola with vertex at (0,40) and passing through (50,30).

We use the formula:

(x − h)² = 4a(y − k)

Now (h, k) is (0, 40) so we have: x² = 4a(y − 40) and the parabola passes through (50, 30), so

(50)² = 4a(30 − 40)

2500 = 4a(−10) and this gives 4a = −250

So the equation of the side of the barrel is

x² = -250(y − 40), that is,

y=−x2250+40

We need to find the volume of the cask which is generated when we rotate this parabola between x = -50 and x = 50 around the x-axis.

Vol=π∫bay2 dx=π∫50−50(−x2250+40)2dx=π∫50−50(x462500−80x2250+1600)dx=π[x5312500−80x3750+1600x]50−50

Now, since

(−50)⁵ = −50⁵,
(−50)³ = −50³, and
(−50) = −50,

we can reduce the amount of writing somewhat and put:

Vol=2π[(50)5312500−80(50)3750+1600(50)]=425162 cm3=425.2 L

So the wine cask will hold 425.2 L

2. Volume of a watermelon

A watermelon has an ellipsoidal shape with major axis 28 cm and minor 
axis 25 cm. Find its volume.

Historical Approach: Before calculus, one way of approximating the 
volume would be to slice the watermelon (say in 2 cm thick slices) and 
add up the volumes of each slice usingV=πr2h.

Interestingly, Archimedes (the one who famously jumped out of his bath 
and ran down the street shouting "Eureka! I've got it") used this approach
 to find volumes of spheres around 200 BC. The technique was almost 
forgotten until the early 1700s when calculus was developed by Newton 
and Leibniz.

We see how to do the problem using both approaches.

Volume using historical method:

Because the melon is symmetrical, we can work out the volume of one half of the melon, 
and then double our answer.
The radii for the slices for one half of a particular watermelon are found from measurement
 to be:
0,6.4,8.7, 10.3,11.3, 12.0,12.4,12.5.
The approximate volume for one half of the melon using slices 2 cm thick would be:
Vhalf=π×[6.42+8.72+10.32+11.32+12.02+12.42+12.52]×2=π×8040.44×2=5054.4



So the volume for the whole watermelon is about
5054.4×2=10109 cm3=10.1 L.

"Exact" Volume (using Integration):

We are told the melon is an ellipsoid. We need to find the equation of the cross-sectional ellipse with major axis 28 cm and minor axis 25 cm.

We use the formula (from the section on ellipses):

x2a2+y2b2=1

where a is half the length of the major axis and b is half the length of the minor axis.

For the volume formula, we will need the expression for y² and it is easier to solve for this now (before substituting our a and b).

x2a2+y2b2=1b2x2+a2y2=a2b2a2y2=a2b2−b2x2=b2(a2−x2)y2=b2a2(a2−x2)

Since a=14 and b=12.5, we have:

y2=12.52142(142−x2)=0.797(196−x2)

NOTE: The a and b that we are using for the ellipse formula are not the same a and b we use in the integration step. They are completely different parts of the problem.

Using this, we can now find the volume using integration. (Once again we find the volume for half and then double it at the end).

Vhalf=π∫140y2dx=π∫1400.797(196−x2)dx=0.797π∫140(196−x2)dx=2.504[196x−x33]140=2.504[196(14)−1433]=2.504×1829.33=4580.65 cm3

So the watermelon's total volume is 2×4580.65=9161 cm3 or 

9.161 L
 This is about the same as what we got by slicing the watermelon and adding the volume of the slices.