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Integrating with substitution methods (part 1)

  1. Dec 8, 2013 #1
    Mod note: Edited the LaTeX so that the exponents show up correctly.
    1. The problem statement, all variables and given/known data

    This is from my Calculus II exam practice papers. We're currently dealing with different substitution methods (whichever apply to the given problem).


    2. Relevant equations

    [itex]
    \int \frac {\sqrt{1 - x^2}} {x^{4}} dx
    [/itex]


    3. The attempt at a solution

    So I started off by attempting to rewrite it as:

    [itex]
    \int (1 - x^2)^\frac {1}{2} (x^{-4}) dx
    [/itex]


    So, let: [itex]u = x^2[/itex] and [itex]du = 2x dx[/itex]


    Rewriting formula again:

    [itex]
    \int (1 - u)^\frac {1}{2} (u^{-2}) du
    [/itex]


    Integrating.....

    [itex] \frac {-2}{3} (1 - u)^\frac {3}{2} (\frac {u^{-1}}{-1}) + constant [/itex]


    Subbing the U out and fixing things up a bit....

    [itex]
    \frac {2}{3}(1 - x^2)^\frac {3}{2} (\frac {1}{x^2}) + constant
    [/itex]


    Quite frankly, I am honestly not sure if I am doing this correctly so far. This is actually the third process I've tried (the first 2 were even longer). I'd like to be shed some light if I am going about this correctly or not. Thanks in advance.


    PS: I posted (part 1) on the topic since I will require assistance in a few other examples in this paper. Should I keep posting my questions regarding the other problems in this same thread or a new one?
     
    Last edited: Dec 8, 2013
  2. jcsd
  3. Dec 8, 2013 #2

    Mark44

    Staff: Mentor

    No, this isn't correct at all. It is not true that ##\int f(x) g(x) dx = \int f(x) dx * \int g(x)dx##.
     
  4. Dec 8, 2013 #3

    Mark44

    Staff: Mentor

    A substitution might work, but then again, it might not. I would be more inclined to go with a trig substitution.
    New thread.
     
  5. Dec 8, 2013 #4
    Tried it without using any substitution. Results don't exactly match Wolfram Alpha (for reference).

    [itex]
    \int (1 - x^2)^\frac {1}{2} (x^-4) dx
    [/itex]

    [itex]
    \frac {2}{3} (1 - x^2)^\frac {3}{2} (\frac {x^-3}{-3}) + constant
    [/itex]

    [itex]
    \frac {2}{3} (1 -x^2)^\frac{3}{2} (\frac {1}{-3x^3}) + constant
    [/itex]


    This is somewhat closer to what W|A gets. Yes, I know it can simplify further but I'd like to know if this is wrong as well anyways.
     
  6. Dec 8, 2013 #5
    Reviewing my notes after remembering something about trigonometric substitutions I found:

    [itex] \sqrt {a^2 - u^2} [/itex], let [itex]u = a \sin\theta[/itex] and [itex]du = a \cos\theta d\theta[/itex]

    Is that what you were referring to, Mark44?
     
  7. Dec 8, 2013 #6

    Mark44

    Staff: Mentor

    In LaTeX, if your exponent is more than one character, you have to put it in braces {}.

    Your integrand should look like this: (1 - x^2)^\frac {1}{2} x^{-4}
    The above is completely wrong for the reason I gave in my previous post. Integration doesn't work this way.
     
  8. Dec 8, 2013 #7

    Mark44

    Staff: Mentor

    Yes.
     
  9. Dec 8, 2013 #8

    Mark44

    Staff: Mentor

    There's another problem. In your first post you started with
    ## \int \frac{\sqrt{1 - x^2}}{x^{-4}}dx##
    and then later you had
    ## \int \sqrt{1 - x^2} x^{-4}dx##
    Those two integrands aren't equal. When you move x-4 from the denominator, it becomes x4 in the numerator.
     
  10. Dec 8, 2013 #9
    That -4 was a typo. Fixed.
     
  11. Dec 8, 2013 #10
    This is what I did afterwards and where I pretty much left the problem at before moving on to other problems in my paper:

    [itex]
    \int \frac {\sqrt{1 - x^2}}{x^4} dx
    [/itex]

    Let u = x, du = 1dx

    [itex]
    \int \frac {\sqrt{1 - u^2}}{u^4} dx
    [/itex]


    Now, let u = sin[itex]\theta[/itex], du = cos[itex]\theta d\theta[/itex]


    So I end up with:

    [itex]
    \int \frac {sin^2 \theta cos\theta}{sin^4 \theta} d\theta
    [/itex]


    Which becomes:

    [itex]
    \int \frac {cos\theta}{sin^2 \theta} d\theta
    [/itex]


    Which W/A says is equal to [itex]-csc(x) + constant[/itex] for some reason.

    And I think I just figured it out. Give me a few minutes. Latex takes forever to write.
     
  12. Dec 8, 2013 #11
    Okay, so... I break it apart. It being:

    [itex]
    \int \frac {cos\theta}{sin^2 \theta} d\theta
    [/itex]


    Which pretty much becomes this:

    [itex]
    \int (\frac {1}{sin \theta}) (\frac {cos \theta}{sin \theta} d\theta
    [/itex]

    Applying identities, yada yada yada...

    [itex]
    \int csc\theta cot\theta d\theta
    [/itex]


    Which can be easily integrated into something that makes bloody sense.

    [itex]
    -csc\theta + constant
    [/itex]

    At this point, we're supposed to use the reference triangle (or however it's called in English, triangulo de referencia for us) and do some voodoo with the drawing.

    So, cosecant of an angle = [itex]\frac{hyp}{opp}[/itex], I am guessing it's [itex]-csc \frac{1}{x} + constant[/itex]?

    That or I replace [itex]-csc \theta[/itex] with the original sin, which would be [itex]-sin(x) + constant[/itex] for the final correct answer. Or something else that I don't see and I give up :p
     
    Last edited: Dec 8, 2013
  13. Dec 8, 2013 #12

    Mark44

    Staff: Mentor

    Now, undo your substitution. Your work to this point looks reasonable, but I didn't check it closely.

    The trig substitution was u = sin(θ), so θ = sin-1(u).
    -csc(θ) = -1/sin(θ) = -1/u = -1/x, which can't be right.

    You did a substitution a couple of posts ago: Let u = x. You should never do this substitution, because it's useless. All you're doing with this substitution is changing the letters in your problem, without doing anything useful.

    I'll take a closer look at your work to see where you're going wrong.
     
  14. Dec 8, 2013 #13

    Mark44

    Staff: Mentor

    As already mentioned, u = x is a useless substitution, and is one that never does any good. Let's go back to the original integral:
    [itex]\int \frac {\sqrt{1 - x^2}}{x^4} dx[/itex]

    ##x = sin(\theta), dx = cos(\theta) d\theta##

    This is OK.

    No.
    From your reference triangle, which you should have used before you made your trig substitution, you would have cos(θ) = √(1 - x2), so the integral above should be
    $$ \int \frac{cos^2(θ)dθ}{sin^4(θ)}$$
    Use the reference triangle before you make the substitution.
     
  15. Dec 9, 2013 #14
    I give up on this one then. Thanks for your help though.

    Professor always uses the triangle right at the end though.
     
  16. Dec 9, 2013 #15

    Mark44

    Staff: Mentor

    IMO, that's kind of dumb to wait until the end. I use it at the beginning, as I don't want to clutter up my mind with which formula goes with which substitution. I draw the triangle and label it according to what's in the radical. From the triangle you can get every relationship in the original integral.
     
  17. Dec 9, 2013 #16

    Mark44

    Staff: Mentor

    Why give up? You were nearly done.
     
  18. Dec 9, 2013 #17
    Never give up man, this isn't that difficult. I know I'm not supposed to give you answers, but I hate when people give up.
    $$\int \frac{1-x^{2}}{x^{4}}dx$$
    Use ##x=sin\theta## and ##dx=cos\theta d\theta##
    Now you end up with:
    $$\int \frac{\sqrt{1-sin^{2}\theta}}{sin^{4}\theta} cos\theta d\theta$$
    Since ##1-sin^{2}\theta=cos^{2}\theta## substitute it into the above and you'll see you end up with:
    $$\int \frac{\sqrt{cos^{2}\theta}}{sin^{4}\theta}cos\theta d\theta$$
    You should see that the square root will end up with just cosθ multiplied together you get:
    $$\int \frac{cos^{2}\theta}{sin^{4}\theta}d\theta$$
    Now finally, using trig identities:
    $$\int cot^{2}\theta csc^{2}\theta d\theta$$
    You should be able to solve it from there, use a simple u-substitution.

    That's the basic format of every one of these problems. Use a trig substition to get a square root of a squared trig function. Then the square root will get rid of the square and you'll be able to solve using trig identities.
     
  19. Dec 9, 2013 #18
    Sorry, but I really just don't get it. Gave up on it because I have quite a few more problems to solve in my papers. I don't know if I wrote down something wrong or if the professor made a mistake at some point.

    The table I have says to let [itex]u = a * sin \theta[/itex] while [itex]du = a * cos \theta d\theta[/itex].

    Somehow we get [itex]\sqrt{a^2 - a^2 * sin^2 \theta}[/itex] which equals [itex]\sqrt{a * cos \theta}[/itex].

    I honestly have no idea where I messed up combining that information from the table into my process.
     
  20. Dec 9, 2013 #19

    Mark44

    Staff: Mentor

    If you draw the triangle first, you don't have to rely on memorizing formulas, especially ones that you don't understand. Because the radical is √(1 - x2), label the hypotenuse of your right triangle as 1, and the side opposite angle θ as x. From that you have sin(θ) = x/1. That's where the substitution x = sin(θ) comes from, and where dx = cos(θ) dθ comes from. It's also easy to see that cos(θ) = ##\frac{\sqrt{1 - x^2}} {1}.##

    Now you're ready to replace x and dx with expressions that involve θ and dθ.
    In this problem a = 1, so the first radical is ##\sqrt{1 - sin^2(θ)}##. That's equal to ##\sqrt{cos^2(θ)}##, which in turn is equal to cos(θ). As long as θ is between 0 and ##\pi/2## (which is true in a right triangle), you don't have to consider the negative possibility for ##\sqrt{cos^2(θ)}##.
     
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