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Trig substitution help (easy one !)

  1. Apr 3, 2005 #1
    Use trig substitution to find [tex]\int_{0}^{5} \frac{dt}{25 + x^2}dt[/tex]

    I can solve it to here [tex]\int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{(25 + tan^2\theta)^2}[/tex]

    and from this point i can factor the denominator into [tex]{625(1+ \tan^2\theta)}^2[/tex]

    which becomes [tex]625\sec^4\theta[/tex]

    now i have the integral [tex]\int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{625\sec^4\theta}[/tex]

    this now reduces to [tex]\int_{0}^{\frac{\pi}{4}}\frac{cos^2\theta}{25}[/tex]

    and at this point i can use a power reducing formula to get rid of the [tex]\cos^2\theta[/tex]

    assuming that the last integral is correct and that i use the power reducing formula to reduce [tex]\cos^2\theta[/tex] correctly, what am i doing wrong?
    i have a TI-89 graphing calculator, and when i integrat this problem on it i get a different answer than when i do it by hand. where is my mistake?

    this post is incorrect look further down for the correction.
    i'm really sorry about this.
    Last edited: Apr 3, 2005
  2. jcsd
  3. Apr 3, 2005 #2
    You can use the double angle formula for cos 2x to get rid of cos^2 x. cos 2x = 2cos^2 x - 1.
  4. Apr 3, 2005 #3
    As the integral stands now, you could use another formula to find the answer, but in order to put the integral into a form where you can use a trig substution you have to play around with it:

    [tex]25 + t^2 = (\sqrt{25 + t^2})^2[/tex]

    So the integral becomes:

    [tex]\int_{0}^{5}\frac{dt}{(\sqrt{25 + t^2})^2}[/tex]

    Make the substutions:
    [tex]t = 5 \tan{\theta}[/tex]
    [tex]dt = 5 \sec^2{\theta} d\theta[/tex]

    and work it from there. :)
    Last edited: Apr 3, 2005
  5. Apr 3, 2005 #4
    This is actually a standard integral. Anyways, you seem to have made a few errors. Here we want

    [tex] \int_0^5 \frac{dt}{25 + t^2}[/tex]

    let [itex] t = 5\tan{\theta} \Longrightarrow dt = 5 \sec^2{\theta}[/itex] and note that [itex]t=5 \Longrightarrow 5 = 5\tan{\theta} \Longrightarrow \theta = \pi / 4[/itex]. Thus we just have

    [tex] \int_0^{\frac{\pi}{4}} \frac{5\sec^2 \theta}{25 + (5 \tan \theta)^2} \ d\theta[/tex]

    [tex] = \frac{1}{5}\int_0^{\frac{\pi}{4}} \frac{\sec^2 \theta}{1 + \tan^2 \theta} \ d\theta[/tex]

    now, review your trig identities and see if you can find an easy way to do this :smile:

    The standard integral is

    [tex]\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) \ + \ C,[/tex]

    by the way.
    Last edited: Apr 3, 2005
  6. Apr 3, 2005 #5

    I made a huge mistake in my original post!!!!!! the first integral is

    [tex]\int_{0}^{5} \frac{dt}{(25 + x^2)^2}dt[/tex]

    and what follows is what i get. i'm really sorry about posting this problem incorrectly.
  7. Apr 3, 2005 #6
    i'm putting this back to the top because i screwed it up the first time.
  8. Apr 3, 2005 #7
    No problem :smile:

    Anyways, in that case your work is almost right. We want

    [tex] \int_0^5 \frac{dx}{(25+x^2)^2}[/tex]

    substitute [itex]x = 5\tan \theta \Longrightarrow dx = 5\sec^2 \theta \ d\theta[/itex] and so [itex]x=5 \Longrightarrow \theta = \pi / 4[/itex]. Thus our integral is the same as

    [tex] \int_0^{\frac{\pi}{4}} \frac{5\sec^2 \theta}{(25 + 25\tan^2 \theta)^2} \ d\theta[/tex]

    [tex] = \frac{1}{125}\int_0^{\frac{\pi}{4}} \frac{\sec^2 \theta}{\sec^4 \theta} \ d\theta[/tex]

    [tex] = \frac{1}{125}\int_0^{\frac{\pi}{4}} \cos^2 \theta \ d\theta[/tex]
  9. Apr 3, 2005 #8
    i worked this down to exactly the same thing that you did but when i use my
    TI-89 graphing calculator to integrate this i get a different answer than what i get by hand.
  10. Apr 3, 2005 #9
    [tex]\frac{1}{125}\int_0^{\frac{\pi}{4}} \cos^2 \theta d\theta[/tex]

    [tex] = \frac{1}{125} \int_0^{\frac{\pi}{4}} \frac{1}{2} \left(1 + \cos (2\theta)) \ d\theta[/tex]

    [tex] = \frac{1}{250}\left[\theta + \frac{\sin (2\theta)}{2} \right]_0^{\frac{\pi}{4}}[/tex]

    [tex] = \frac{1}{250}\left(\frac{\pi}{4} + \frac{1}{2}\right).[/tex]
  11. Apr 3, 2005 #10
    now i see my mistake. thankyou Data
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