Page 2 of 3 FirstFirst 123 LastLast
Results 16 to 30 of 39

Math Help - even more trig questions!

  1. #16
    Junior Member
    Joined
    Jun 2010
    From
    United Kingdom
    Posts
    64
    Quote Originally Posted by Ackbeet View Post
    So, number 4 works like this. You've got your integral,

    \int_{0}^{\sqrt{3}}\ln\left(\frac{2}{\sqrt{4-x^{2}}}\right)dx.

    Now then, suppose you draw your triangle with theta as the lower left angle. The squared terms have opposite signs, so you let 2 be the hypotenuse, and x the opposite side to theta. That means the adjacent side has length \sqrt{4-x^{2}}. You need to transform the limits, the integrand, and the differential so that there are no more x's, only theta's. Here are some relevant equations that let you do that:

    \frac{x}{2}=\sin(\theta)

    \frac{2}{\sqrt{4-x^{2}}}=\sec(\theta).

    So, hopefully, getting the integrand to have only theta's should be fairly straight-forward. How do you propose to get the limits and the differential to have only theta's?

    Can I replace \frac{2}{\sqrt{4-x^{2}}} found in the integral with sec(theta) such that I am integrating ln(sec(theta))?

    In terms of the limits, they have no xs in them so can they be left as they are?
    Last edited by maggiec; June 26th 2010 at 03:44 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Yes, you can do the replacement with \sec(\theta). No, you cannot leave the limits as they are. They are values of x, because of the differential. You have to change them to values of \theta. Or, if you want, you can convert the antiderivative back into x's before you evaluate at the limits. I generally find the first alternative less work, but you can do as you wish.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Junior Member
    Joined
    Jun 2010
    From
    United Kingdom
    Posts
    64
    Quote Originally Posted by Ackbeet View Post
    Yes, you can do the replacement with \sec(\theta). No, you cannot leave the limits as they are. They are values of x, because of the differential. You have to change them to values of \theta. Or, if you want, you can convert the antiderivative back into x's before you evaluate at the limits. I generally find the first alternative less work, but you can do as you wish.
    Thus the limits change:

    when x=√3, sin \theta = (√3)/2 hence \theta = 60 degrees
    when x = 0, \theta = 0

    so the new limits are 0 and 60?

    any hints on how to integrate ln(sectheta)?
    Follow Math Help Forum on Facebook and Google+

  4. #19
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Your limits look correct to me. But \ln(\sec(\theta)) isn't all that's in the integrand, if I'm not mistaken. What does the new differential look like?
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Junior Member
    Joined
    Jun 2010
    From
    United Kingdom
    Posts
    64
    Quote Originally Posted by Ackbeet View Post
    Your limits look correct to me. But \ln(\sec(\theta)) isn't all that's in the integrand, if I'm not mistaken. What does the new differential look like?
    I thought it was y= \ln\left(\frac{2}{\sqrt{4-x^{2}}}\right)

    but then sec(theta)= \frac{2}{\sqrt{4-x^{2}}}\right)

    thus the differential becomes \ln(\sec(\theta))?
    Follow Math Help Forum on Facebook and Google+

  6. #21
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    No, that's the integrand. You've successfully put the integrand into thetas. What you have to do now is find the differential dx. This is the same thing you have to do with any substitution (u substitution or whatever). In order to find the differential, I would recommend finding the simplest relation possible between x and theta from the triangle you drew. What do you get?
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Junior Member
    Joined
    Jun 2010
    From
    United Kingdom
    Posts
    64
    Quote Originally Posted by Ackbeet View Post
    No, that's the integrand. You've successfully put the integrand into thetas. What you have to do now is find the differential dx. This is the same thing you have to do with any substitution (u substitution or whatever). In order to find the differential, I would recommend finding the simplest relation possible between x and theta from the triangle you drew. What do you get?

    \sin(\theta)=\frac{x}{2}

    Oh I have to find dx/d(theta)..
    Follow Math Help Forum on Facebook and Google+

  8. #23
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Excellent. Now take the differentials of both sides to get the needed relation.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    Junior Member
    Joined
    Jun 2010
    From
    United Kingdom
    Posts
    64
    2sin(theta) = x
    hence dx/d(theta) = 2 cos(theta)

    hence the integral becomes

    integral of: 2ln(sec(theta)).cos(theta).d(theta) between 0 and 60 degrees

    Now integration by parts?
    Follow Math Help Forum on Facebook and Google+

  10. #25
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    That sounds like a plan. Try it!
    Follow Math Help Forum on Facebook and Google+

  11. #26
    Junior Member
    Joined
    Jun 2010
    From
    United Kingdom
    Posts
    64
    Quote Originally Posted by Ackbeet View Post
    That sounds like a plan. Try it!
    Can I just check, differentiating ln(sec(theta)) you get [sec(theta).tan(theta)]/(e^u) ?
    Last edited by maggiec; June 27th 2010 at 03:46 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Junior Member
    Joined
    Jun 2010
    From
    United Kingdom
    Posts
    64
    I've sort of gotten there but half way... theres a final term I lost somewhere but Im not sure where!
    Here's what I've done, hope it's clear to see!
    even more trig questions!-screen-shot-2010-06-27-11.46.22.jpg
    Follow Math Help Forum on Facebook and Google+

  13. #28
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    I don't agree with your differentiation of \ln(\sec(\theta)). Use the Chain Rule. What do you get?
    Follow Math Help Forum on Facebook and Google+

  14. #29
    Junior Member
    Joined
    Jun 2010
    From
    United Kingdom
    Posts
    64
    Quote Originally Posted by Ackbeet View Post
    I don't agree with your differentiation of \ln(\sec(\theta)). Use the Chain Rule. What do you get?
    I used that exact same method earlier on in the question and got the correct answer for the gradient?

    I thought I couldn't use the chain rule because u isn't a constant..

    but using the chain rule I get: du/d(theta) = tan(theta)
    Follow Math Help Forum on Facebook and Google+

  15. #30
    Junior Member
    Joined
    Jun 2010
    From
    United Kingdom
    Posts
    64
    I still get sin(theta)tan(theta) having to be integrated thus giving me the same result, so both differentials lead me to the same overall integral, not sure why..
    Follow Math Help Forum on Facebook and Google+

Page 2 of 3 FirstFirst 123 LastLast

Similar Math Help Forum Discussions

  1. Few Trig Questions
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 20th 2010, 07:14 AM
  2. Find the value of 'trig of inverse trig' questions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 9th 2010, 06:37 PM
  3. Replies: 6
    Last Post: November 20th 2009, 05:27 PM
  4. 2 trig Questions
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: January 19th 2009, 11:23 PM
  5. Trig Questions !!
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: October 20th 2008, 12:34 PM

Search Tags


/mathhelpforum @mathhelpforum