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Trigonometric Substitution

  1. Sep 20, 2007 #1
    [tex]\int\sqrt{16-(2x)^{4}}xdx[/tex]

    Hint says you may like to use the identity sin(theta)cos(theta)= sin(2theta)/2

    However, I think I found a way to use 1-sin^2(theta)=cos^2(theta)

    First, (2x)^4 = 16x^4

    So make it 16(1-x^2)^2.

    Take the 16 out of the root and the integral and you have:

    [tex]4\int\sqrt{1-(x^{2})^{2}}xdx[/tex]

    Sub in sin(theta) for x^2. Sub in cos^2(theta) for 1-sin^2(theta). Sub in [tex]\sqrt{sin(\vartheta)[/tex] for the x outside the root and [tex]\frac{1}{2\sqrt{sin(\vartheta)}}[/tex] for dx.

    Cancel out the sqrt{sin} on the top and bottom and drag the 1/2 outside the integral. Anti-derivative of cos is sin and you have 2sin(theta). Sin(theta) = x^2.

    And so the final answer is; [tex]2x^{2}+c[/tex]

    Correct?
     
  2. jcsd
  3. Sep 20, 2007 #2

    Dick

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    Is the derivative of 2x^2+c the same as the original integrand? I don't think so.
     
  4. Sep 20, 2007 #3
    Hrmm....could you point me in the right direction?
     
  5. Sep 20, 2007 #4

    Dick

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    Your substitution sin(theta)=x^2 is a great start. So cos(theta)*dtheta=2*x*dx. See where that takes you. I promise that's heading in the right direction.
     
  6. Sep 20, 2007 #5
    Still a little stumped.

    Realized I missed the chain rule in 2 steps in my original post...damn rustiness.

    Got to [tex]\frac{-sin\vartheta(cos\vartheta)^{3}}{3}[/tex]

    First and foremost, am I correct to this point? Then, I can't figure out the back substitutions here.
     
  7. Sep 20, 2007 #6

    Dick

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    I get to integral of cos(theta)^2.
     
  8. Sep 20, 2007 #7
    Yea, I posted the anti-derivative of cos(theta)^2. I'm at the point where you need to substitute in x's for the trig functions.
     
  9. Sep 20, 2007 #8

    Dick

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    That is NOT the antiderivative of cos(theta)^2. To do the antiderivative of cos(theta)^2 use that cos(2*theta)=cos(theta)^2-sin(theta)^2 (from the addition relation for cos) and that sin(theta)^2=1-cos(theta)^2. Solve for cos(theta)^2 in terms of cos(2*theta). Then it's easy to integrate cos(2*theta). After the integration is done you'll find that the hint actually does come in handy for expressing the result in terms of x.
     
  10. Sep 20, 2007 #9

    Dick

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    Hint: When you think something is the antiderivative of something else then try differentiating it to check. Could save you some time, since you seem to be getting antiderivatives wrong a lot.
     
  11. Sep 20, 2007 #10

    Dick

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    You could also try checking an integral table or an online site that computes integrals to check yourself. You can't do that on an exam, but it could be handy to avoid wasting time while practicing with homework problems.
     
  12. Sep 20, 2007 #11
    Definitely having problems with the little/old things. That year off killed me. Just have to do all the little things a few times before it comes back to me.

    Anyways. Came up with 2cos^2(theta)=cos2theta + 1

    So integral cos2theta + 1

    Comes to Sintheta * costheta + theta

    Don't know what to do with that theta at the end. So I'm guessing I'm wrong.
     
  13. Sep 20, 2007 #12

    Dick

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    You aren't that wrong this time. theta=arcsin(x^2). And what does sin(theta)*cos(theta) become? Hammer the multiplicative constants down and I think you've got it.
     
  14. Sep 20, 2007 #13
    Just checking to make sure I'm not aiming for the wrong thing;

    I'm currently at sin(2theta)/2 + arcsin(x^2)+c

    An integration website tells me that I should have [tex]\sqrt{1-x^{4}}x^{2}[/tex]+ arcsin(x^2)+c

    Was the site right?
     
  15. Sep 20, 2007 #14

    Dick

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    It's time to use the hint for sin(2theta)! You and the site are both right.
     
    Last edited: Sep 20, 2007
  16. Sep 20, 2007 #15
    Okay, so I got to the answer that the site gave me. But question, wasn't that 4 brought outside the integral used while converting cos^2theta to cos2theta and when solving for dx?
     
  17. Sep 20, 2007 #16

    Dick

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    Probably. Sorry, I'm fading a bit and didn't want to check it myself. Just wanted to make sure you did.
     
  18. Sep 20, 2007 #17
    Alright, thanks. =P

    I owe ya, but I imagine there isn't much math I could help ya with.
     
  19. Sep 20, 2007 #18

    Dick

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    You help me by making me feel that my time was not wasted on this internet busyness. The quicker you catch on, the better I feel. You're getting better, right?
     
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