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Techniques of Integration [need feedback]

  1. Aug 12, 2009 #1
    {I wrote this a while ago, thought i would share. I know it's not complete, far from complete actually, but if there's enough interest I'll finish it in notime...heak i might even write more expand the index.}

    1)Direct Use of Formula
    2)Numerical Integration
    3)Substitution Method
    4)Integration by Parts
    5)Trigonometric Integrals
    6)Trigonometric Substitution
    7)Integrals of Hyperbolic and Inverse hyperbolic Functions
    8)The Method of Partial Fractions
    9)Improper Integrals

    {I plan to tweak this list a little}

  2. jcsd
  3. Aug 12, 2009 #2
    1) Direct Use of Formula
    This is something you should know from your previous Calculus class. Here is a quick run through some of the basic integration formulas that you are expected to know before going any further:

    [tex]1) \int u^n du = \frac{u^{n+1}}{n+1} + C; \hspace{1cm} \text{provided} \hspace{1cm} n\neq -1[/tex]

    [tex]2) \int \frac{du}{u} = \ln|u| + C[/tex]

    [tex]3) \int e^u du = e^u + C[/tex]

    [tex]4) \int a^u du = \frac{a^u}{\ln(a)} + C[/tex]

    [tex]5) \int \sin (u) du = - \cos(u) + C[/tex]

    [tex]6) \int \cos (u) du = \sin (u) + C[/tex]

    [tex]7) \int \sec^2 (u) du = \tan (u) + C[/tex]

    [tex]8) \int \csc^2 (u) du = - \cot (u) + C[/tex]

    [tex]9) \int \tan (u) du = - \ln |\cos (u)| + C[/tex]

    [tex]10) \int \csc (u) \cot (u) du = - \csc (u) + C[/tex]

    [tex]11) \int \tan (u) du = \ln|\sec (u)| + C[/tex]

    [tex]12) \int \cot (u) du = \ln|\sin (u)| + C[/tex]

    [tex]13) \int \sec (u) du = \ln|\sec (u) + \tan(u)| + C[/tex]

    [tex]14) \int \csc (u) du = \ln|\csc (u) - \cot (u)| + C[/tex]

    Practice Set 1: Use Formulas straight up!

    [tex]1) \int 4/x dx[/tex]

    [tex]2)\int 3^x dx[/tex]

    [tex]3)\int (\sec(x) + \csc^2 (x)) dx[/tex]

    Practice Set 2: Algebraically manipulate first to match the formula, then use it!

    [tex]1) \int (x + 3)^2 x^5 dx[/tex]

    [tex]2)\int 3x^3-2x+5/(x+3) dx[/tex]

    [tex]3)\int \sec(x) (\sec(x) + \tan(x)) dx[/tex]

    Last edited by a moderator: Aug 19, 2009
  4. Aug 12, 2009 #3
    I think it is best to kick off various integration techniques with Numerical Integration. For one thing, I believe it will help you picturize integration better. More importantly, this method always works, even when many other listed above fail.

    For instance,
    [tex]\int sin(x^2) dx[/tex]
    you can not integrate it with any other listed techniques but this one.


    You should be familiar with these techniques from your previous class, likely previous calculus class. Nevertheless, here are the formulas to refresh your memories.
    [tex] L_N = \triangle x (y_0 + y_1 + ... + y_{N-1} )[/tex]

    [tex] R_N = \triangle x (y_1 + ... + y_N)[/tex]

    [tex]M_N = \triangle x [ f (x_0 + x_1) / 2 + f (x_1 + x_2) / 2 + ... + f (x_{N-1} + x_N) / 2) [/tex]

    2.2 Trapezoidal Rule

    [tex]Area of Trapezoid = 1/2 h (b_1 + b_2) [/tex]]

    where h is the height of the trapezoid and b_1 & b_2 are parallel sides of the trapezoid.

    [tex]T_N = \triangle x/2 (y_0 + 2y_1 + ... + 2y_{N-1} + y_N)[/tex]

    2.3 Simpson's Rule
    [tex] S_N = \triangle x/3 (y_0 + 4y_1 + 2y_2 + 4y_3 + ... + 2y_{N-2} + 4y_{N-1} + y_N)[/tex]

    [tex]S_N = 1/2 (T_{N/2} + 2 M _ {N/2}[/tex]

    Note: Simpson's rule is only valid for [tex]N[/tex] is an even number. Why? Notice the [tex]N/2[/tex] in the second formula.

    2.4 Accuracy and Error

    So by now there should be an obvious question poping in your head: which method is more accurate?

    The answer lies behind the nature of function you're given with.

    For instance, if you are given a function of the form

    [tex] f(x) = C[/tex]
    left, right, or the midpoint method would be accurate (and so would other methods btw). This is b/c the left, right, or the midpoint method divides the area under a given function into small rectangles and if a the given function is parallel to the axes, all area under the function will be occupied by the rectangles.

    However, once you go to a function of the form

    [tex] f(x) = bx + c (line) [/tex]

    the accuracy for left, right, and the midpoint method decreases. Yet, the trapezoidal rule is accurate (and so will the simpsons but why hassle with simpson's method when trapezoidal does the job, right?) This is b/c the trapezoidal rule divides the area under the given function with trapezoids (duh?).

    Next, if you go to function of the form

    [tex] f(x) = ax^2 + bx + c (quadratic)[/tex]

    the accuracy for trapezoidal rule (along with left, right and the midpoint method) decreases. Yet the Simpson's method maintains it's accuracy.

    Now, if you go have a function with power greater than 2 then you'll have to enhance simpson's rule further.

    Example: Evaluate the given integral from 0 (lower boundary) to 2 (upper boundary)

    [tex]\int x^2 dx = x^3/3[/tex]

    [tex]=> 8/3[/tex]

    Now calculate the value using various techniques, and their respective error percentage:

    [tex]L_2 Error = 8/3 - 1 => Error % = 50%[/tex]
    [tex]R_2 Error = 8/3 - 5 => Error % = 87.5 %[/tex]
    [tex]M_2 Error = 8/3 - 7/3 => Error % = 6.25 %[/tex]
    [tex]T_2 Error = 8/3 - 3 => Error % = 12.5 %[/tex]
    try out Simpson's rule yourself.

    [tex]Error (T_N) \leq k_2 (b-a)^3/{12N^2}[/tex]
    [tex]Error (M_N) \leq k_2 (b-a)^3/{24N^2}[/tex]
    [tex]Error (T_N) \leq k_4 (b-a)^5/{180N^4}[/tex]

    {will likely add up few pictures for illustration}
    Last edited by a moderator: Oct 16, 2010
  5. Aug 12, 2009 #4
    3) U-Substitution Method

    It is one of the most easy yet efficient and effective method of integration. It states:

    If [tex]F^\prime(x) = f(x)[/tex] then,

    [tex]\int f(u(x)) u^\prime(x) dx = F(u(x)) + C[/tex]

    Basically, if the integrant has a function, u(x), and it's derivative, u^\prime(x), then use this method.

    For ease of understanding I'll divide this method into two parts namely: U-Substitution on Indefinite Integrals and U-Substitution on Definite Integrals.

    3.1 Substitution Method for Indefinite Integrals

    Recall that Indefinite Integrals are the one without upper and lower boundaries. Lack of boundaries makes it easier.

    Example 3.1.1:

    [tex]\int 2x cos(x^2) dx[/tex]

    {Notice: [tex]2x[tex] is the derivative of [tex]x^2[/tex]}

    Step 1: Make a substitution:

    let [tex]u = x^2[/tex]

    so, [tex]du/dx = 2x => du = 2x dx[/tex]

    Step 2: Write the integral in terms of u and evaluate:

    [tex]\int 2x cos(x^2) dx = \int cos(u) du[/tex]

    [tex] = sin(u) + C[/tex]

    Example 3.1.2:

    [tex]\int (2x+1) e^{x^2+x} dx[/tex]

    {Notice: [tex]2x+1[/tex] is the derivative of [tex]x^2+x[/tex]}

    Step 1: Make a substitution:

    let [tex]u = x^2 + x [/tex]

    so [tex]du/dx = 2x + 1 => du = (2x+1) dx[/tex]

    Step 2: Write the integral in terms of u and evaluate:

    [tex]\int (2x+1) e^{x^2+x} dx = \int e^u du[/tex]

    [tex]= e^u + C[/tex]

    Example 3.1.3:

    [tex]\int (ln x)^2/x dx[/tex]

    {Notice [tex]1/x[/tex] is the derivative of [tex]lnx[/tex]}

    Step 1: Make a substitution:

    let [tex]u= lnx[/tex]

    so [tex] du/dx = 1/x => du = 1/x dx[/tex]

    Step 2: Write the integral in terms of u and evaluate:

    [tex]\int (ln x)^2/x dx = \int 1/ \sqrt u du[/tex]

    [tex]= \int u ^{-1/2}[/tex]

    [tex]= 2u^{1/2} + C[/tex]

    Example 3.1.4:

    [tex]\int 1/\sqrt {x-1} dx[tex]

    {Notice: [tex]1[/tex] is the derivative of [tex] x-1[/tex]}

    Step 1: Make a substitution:

    let [tex]u = x-1[/tex]

    so [tex]du/dx = 1 => du = dx[/tex]

    Step 2: Write the integral in terms of u and evaluate:
    [tex]\int 1/\sqrt {x-1} dx = \int 1/\sqrt u du[/tex]

    [tex]=\int u^{-1/2} du = 2u^{1/2} + C[/tex]

    Example 3.1.5

    [tex]\int cot x dx[/tex]

    {Notice: [tex]cot x = cos x / {sin x}[tex] and [tex]cos x[/tex] is really a derivative of [tex]sin x[/tex]}

    Step 1:

    let [tex]u = sin x[/tex]

    so [tex]du = cos x dx[/tex]

    Step 2:

    [tex]\int cosx/ {sin x} dx = \int 1/u du[/tex]

    [tex]= ln|u| + C[/tex]

    Exercise 3.6: Try this yourself!

    [tex]a) \int x sin (x^2) dx[/tex]

    Ans: [tex]-1/2 cos(u) + C[/tex]

    [tex]b) \int tan x dx[/tex]

    Ans: [tex]- ln|sinx| + C[/tex]

    [tex]c) \int x^2 \sqrt{4x-1} dx[/tex]
    Last edited: Aug 12, 2009
  6. Aug 12, 2009 #5
    3.2 Substitution Method for Definite Integrals

    Recall that Definite Integrals are those with lower and upper boundaries i.e. the integral is defined.

    [tex]\int f(u(x)) u^\prime(x) = \int f(u) du[/tex]

    While substitution method on definite integrals it is very important that you keep track of your boundaries. If you do not wish to keep track of your boundaries what you can do is solve given definite integral as an indefinite integral (i.e. disregard the boundaries), use u-substitution, find the anti-derivative, and then plug in the boundaries (both method are shown in example 3.2.1 below).
    Example 3.2.1
    [tex]\int 1/\sqrt{x-1} dx[/tex]

    {Notice: [tex]1[tex] is the derivative of [tex]x-1[/tex]}

    METHOD 2 - Not keeping track of boundaries

    Step 1: Make a substitution:

    let u = x-1
    so du = dx

    Step 2: Write the integral in terms of u and evaluate:

    [tex]\int 1/\sqrt{x-1} dx = \int 1/\sqrt u du [/tex]

    [tex]= 2 u ^ {1/2}[/tex]

    Step 3: Write the result in terms of x:

    [tex]2 u ^ {1/2} = 2 (x-1) ^{1/2}[/tex]

    Step 4: Plug boundaries in anti-derivative found in step 3

    [tex]2 (5-1) - 2 (2-1) = 4 - 2 = 2[/tex]

    METHOD 1 - Keeping track of boundaries

    Step 1: Make a substitution and find new boundaries w.r.t u

    let u = x-1

    so du = dx

    we know, u(x) = x-1 (assumption above)

    so evaluating u(x) at boundaries we get

    u(2) = 2-1 = 1

    u(5) = 5-1 = 4

    Step 2: Write the integral (integrant as well as the boundaries) in terms of u and evaluate:

    [tex]\int 1/\sqrt{x-1} dx[tex]

    [tex]= \int 1/\sqrt u du[/tex]

    [tex]= 2 (4)^{1/2} - 2(1)^{1/2} = 4-2 = 2 [/tex]
    Last edited: Aug 12, 2009
  7. Aug 12, 2009 #6
    4) Integration by Parts

    Recall: The Product Rule

    [tex] d/dx (u(x) * v(x)) = u(x) * v'(x) + v(x) * u'(x)[/tex]

    Integrating both sides, we have:

    [tex]u(x) + v(x) = \int u(x) . v'(x) dx + \int v(x) . u'(x) dx[/tex]
    Integration by parts is helpful when you are trying to integrate product of two functions. One of the function you assume to be u(x) and other to be v'(x) and apply the formula above.

    Then the obvious question arises, how do i know which one is u(x) and which one is v'(x)?

    The brainless way is to randomly pick one to be u(x) and another to be v'(x) but this might work and might not work. If it does not you try it the other way.

    Another way is to think before you execute. Since the formula requires you to find u' (i.e. differentiate u) and v (i.e. integrate v') pick your u such that you can differentiate it and pick your v' such that you can integrate it.

    If that's a lot for you there's another way to do it. Just remember this:

    L = Logarithm
    I = Inverse trig
    A = Algebraic
    T = Trigonometric
    E = Exponential

    {ILATE works too btw}

    whichever function is higher on the list, make it u and the other one is v'.


    [tex]\int xe^x dx[/tex]

    Step 1: Choose u(x) and v'(x)

    X is algebraic and e^x is exponential, since algebraic is higher in the list make x your u(x) and e^x your v'(x)

    Step 2: Compute u'(x) and v(x)

    u'(x) = dx and v(x) = e^x

    Step 3: Use the formula

    [tex] \int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C[/tex]

    {to be continued...}
    Last edited by a moderator: Aug 23, 2009
  8. Sep 4, 2009 #7
    Integration of the series expansion and contour integration are pretty useful. Laplace transforms too. There's also the Howitzer Cannon of integration techniques: differentiation with respect to a parameter.
  9. Mar 3, 2010 #8
    Rubrix, you've covered all of the basics. Very nice.

    The techniques that atropos mentioned are most definitely useful, but are a little more advanced than the list that's already compiled.
  10. Dec 8, 2010 #9
  11. Dec 26, 2010 #10
    By using spherical coordinates how can we get the volume of a right circular cylinder with radius a and height h
  12. May 9, 2011 #11
    As we have direct formulae for integration, it is easy to calculate and interesting.
  13. Jun 21, 2011 #12
    You forgot a few techniques .....
    1/ Tabular Integration By Parts
    2/ Complete the Square
    3/ Long Division
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