1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trigonometric integration

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data

    0[itex]\sqrt{∏}[/itex] xsin(##x^2## -1) dx

    Not sure how I should be formatting this, but the square root of pi is 'on top of' the integral, and zero is 'below'. The expression to integrate is [itex]\sqrt{∏}[/itex] xsin(##x^2## -1) dx.

    3. The attempt at a solution

    As sin integrated is -cos, I'm assuming that sin(##x^2## -1) integrated is -cos(##x^2## -1).

    Using that logic, I get the following:

    [-([itex]\frac{1}{2}[/itex]x cos (##x^2##-1)) + c] (square root of pi on top, 0 below).

    [-([itex]\frac{1}{2}[/itex] ∏ cos(∏-1))] + cos(-1) = 0,308 (using radians)

    I feel that the solution is a bit weird, that I should be left out with a simpler expression or an answer like 0 or 1.

    Am I integrating correctly? I fear not.

    Thanks for any input on this problem in advance.
     
  2. jcsd
  3. Nov 11, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Bad assumption. Does the derivative of ##-\cos(x^2-1)## give you back ##\sin(x^2-1)##? Try a u substitution on your integral..
     
  4. Nov 13, 2013 #3

    I understand what you're saying, but I don't know how to do it. The way I understand the substitution method, I can only do it with two products, here I think I have 3. x, sin and ##x^2## -1.

    If I use the substitution method, I want to rewrite ##x^2## -1 as u and get x sin u. What do I do next then? I remember doing something like f(x)g(x)h(x), but I'm not sure how that would work here.

    Thanks for the idea, and I've obviously looked into it. I'm just not sure how to go on from here.
     
    Last edited: Nov 13, 2013
  5. Nov 13, 2013 #4

    FeDeX_LaTeX

    User Avatar
    Gold Member

    No, you only have a product of two things: ##x##, and ##\sin(x^2 - 1)##.

    You're correct that you need to use the substitution ##u = x^2 - 1##. What, then, is ##\frac{du}{dx}##?
     
  6. Nov 13, 2013 #5
    2x?

    I'm not even close to following this logic in terms of integrating my original expression. How am I supposed to think when I look at my expression?

    I'm both stuck and confused, unfortunately.
     
  7. Nov 13, 2013 #6

    Mark44

    Staff: Mentor

    NO! As already mentioned this product consists of two factors: x and sin(x2 - 1).

    It is vital that you understand that an expression such as sin(x2 - 1) does NOT mean sin * (x2 - 1), any more than log 6 means log * 6 or that √3 means √ * 3. I hope that you recognize that the three misinterpretations are meaningless. Both sin and log are functions, not numbers being multiplied.

    You're getting ahead of yourself. If u = x2 - 1, then du/dx = 2x, or equivalently, du = 2x dx.
    You should be thinking of the chain rule, which is really what ordinary substitution is about.
     
  8. Nov 13, 2013 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    First, forget the trig part- that is not what is causing you trouble! Suppose you had [tex]\int (x^2- 1)^5xdx[/tex]. What do you get when you make the substitution [tex]u= x^2- 1[/tex]?
     
  9. Nov 13, 2013 #8
    As I'm on my phone, I can't reply to the above post for some reason. I'll get back to later.

    To this post:

    I get ∫##u^5## = [itex]\frac{##u^6}{6}[/itex].

    Sorry if I made a sloppy mistake here but it should be right.

    Oh, I didn't notice the x there. It confuses me, but I'll have a look at it later using integration by parts.

    Thanks for your help, I'm sorry I'm a little busy right now to do this properly. I'll get back to this in a few hours.
     
  10. Nov 13, 2013 #9

    Mark44

    Staff: Mentor

    You have a mix of itex and ## tags that are causing the above to not render correctly. Here is the fixed version of what you wrote:
    ##∫u^5 = \frac{u^6}{6}##.

    You are omitting something that is very important - the differential. The integral above should be ∫u5 du. You started off with something of this form: ∫f(x) dx. Using substitution, you need to replace the x expressions and dx with some expression in u plus du that is easier to integrate. If you ignore dx and du, you won't be able to perform the integration successfully.
    No, that's a bad move. Both integrals can and should be done using an ordinary substitution, which is one of the easier integration techniques, and one that should be tried before attempting the more complicated techniques (such as integration by parts).
     
  11. Nov 13, 2013 #10
    Ok, I've gone through this once again, and I was clearly rushing through it and I'm sorry for wasting your time. I can ensure you that this won't happen again.

    I think I have a better understanding of it now, although I still encounter difficulties with substituting back. But I won't get ahead of myself this time, so let me show you what I've done, starting from scratch, using your guidelines.

    Ignore the values for now, as I'm not sure how to format them correctly.

    ∫xsin(##x^2##-1) dx

    u = ##x^2##-1

    [itex]\frac{du}{dx}[/itex] = 2x -> du = 2x dx

    Since du = 2x dx I'm assuming I can substitute x for [itex]\frac{1}{2}[/itex] du to get the following:

    ∫[itex]\frac{1}{2}[/itex] sinu du -> [[itex]\frac{cosu^\frac{3}{2}}{3}[/itex]]

    Now, let's take the 'limits' into play again. The upper limit is [itex]\sqrt{∏}[/itex], the lower limit is 0. After my last step, when the expression is integrated and put into [ ], I'm assuming that they're no longer the same. As I use substitution on the expression, I should do something with the value of the limits as well I think.

    I will look into this, but feel free to look through my work now and see if it's better. At least it makes sense to me after taking what you've said into consideration, but I might've overlooked or ignored something.

    Appreciate what you've done so far. Thanks.
     
  12. Nov 13, 2013 #11

    Mark44

    Staff: Mentor

    Since du = 2x dx, you can replace x dx with (1/2)du.

    In your integral, ∫x sin(u) dx
    becomes (1/2)∫sin(u)du.
    No. You're trying to do too many things at one time. It's much simpler than that. What's an antiderivative of sin(u)? When you have that, replace u with x2 - 1, and then evaluate at the two limits of integration.
     
  13. Nov 14, 2013 #12
    Ok, so I have [itex]\frac{1}{2}[/itex] ∫ sin(u) du.

    The antiderivative of sin(u) is - cos(u).

    I now have [itex]\frac{1}{2}[/itex] [ cos(u) ] with upper limit [itex]\sqrt{∏}[/itex], lower limit 0.

    [itex]\frac{1}{2}[/itex] [ cos(##x^2##-1) ]

    Take the whole expression, put [itex]\sqrt{∏}[/itex] in for x, then subtract the whole expression with 0 for x from that.

    [itex]\frac{1}{2}[/itex] [-cos(∏ - 1)] - [-cos(0-1]. How can I simplify this? And is it correct so far?

    I get [itex]\frac{1}{2}[/itex] (cos(∏-1) + cos(-1) = 0.9996...
     
  14. Nov 14, 2013 #13

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You have dropped a minus sign but otherwise is is now correct. Can you find it? You should get about .54 for a answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Trigonometric integration
  1. Trigonometric Integral (Replies: 16)

  2. Trigonometric Integral (Replies: 1)

  3. Trigonometric integral (Replies: 3)

Loading...