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Math Help - Trig Substitution Integral

  1. #1
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    Trig Substitution Integral

    I'm really spinning my wheels on this one:

    \int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt

    Let's just start with the general antiderivative:

    let t = \sec \theta then dt = \sec\theta \tan\theta

    I know that substituting \sec\theta for \sqrt{t^2-1} yields \tan\theta. This is what I get next (general antiderivative):

    \int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}

    Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:

    \int\cos^2 \theta d\theta

    which integrates to \frac{1}{2}\sin^2\theta + C

    So, two questions:

    1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?

    2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to \theta or get my general antiderivative back into t? This is really confusing to me.

    Can anybody help?

    Thanks!
    Last edited by Ackbeet; June 23rd 2011 at 08:50 AM.
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  2. #2
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    Re: Trig Substitution Integral

    Quote Originally Posted by joatmon View Post
    I'm really spinning my wheels on this one:

    \int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt

    Let's just start with the general antiderivative:

    let t = \sec \theta then dt = \sec\theta \tan\theta

    I know that substituting \sec\theta for \sqrt{t^2-1} yields \tan\theta. This is what I get next (general antiderivative):

    \int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}

    Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:

    \int\cos^2 \theta d\theta
    Correct so far. However, your next step is incorrect. Check that again.

    which integrates to \frac{1}{2}\sin^2\theta + C

    So, two questions:

    1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?

    2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to \theta or get my general antiderivative back into t? This is really confusing to me.

    Can anybody help?

    Thanks!
    Once you finish correcting the above, I would draw a right triangle with theta as the angle, and the sides labeled appropriately for your trig substitution. Then you can figure out how to get back to the t domain. Alternatively, you can transform the limits into the theta domain. Either way should work.
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    Re: Trig Substitution Integral

    try the substitution

    (t^2) - 1 = u^2
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  4. #4
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    Re: Trig Substitution Integral

    Quote Originally Posted by joatmon View Post
    I'm really spinning my wheels on this one:

    \int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt

    Let's just start with the general antiderivative:

    let t = \sec \theta then dt = \sec\theta \tan\theta

    I know that substituting \sec\theta for \sqrt{t^2-1} yields \tan\theta. This is what I get next (general antiderivative):

    \int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}

    Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:

    \int\cos^2 \theta d\theta

    which integrates to \frac{1}{2}\sin^2\theta + C

    So, two questions:

    1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?

    2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to \theta or get my general antiderivative back into t? This is really confusing to me.

    Can anybody help?

    Thanks!
    A hyperbolic substitution might be more appropriate here, as \displaystyle \sinh^2{x} \equiv \cosh^2{x} - 1.

    So in your case, make the substitution \displaystyle t = \cosh{x} \implies dt = \sinh{x}\,dx and the integral \displaystyle \int{\frac{dt}{t^3\sqrt{t^2 - 1}}} becomes

    \displaystyle \begin{align*} \int{\frac{\sinh{x}\,dx}{\cosh^3{x}\sqrt{\cosh^2{x  } - 1}}} &= \int{\frac{\sinh{x}\,dx}{\cosh^3{x}\sinh{x}}} \\ &= \int{\frac{dx}{\cosh^3{x}}} \\ &= \int{\frac{\cosh{x}\,dx}{\cosh^4{x}}} \\ &= \int{\frac{\cosh{x}\,dx}{(1 + \sinh^2{x})^2}} \\ &= \int{\frac{du}{(1 + u)^2}}\textrm{ after making the substitution }u = \sinh{x} \implies du = \cosh{x}\,dx \\ &= -\frac{1}{1 + u} + C \\ &= -\frac{1}{1 + \sinh{x}} + C \\ &= -\frac{1}{1 + \sqrt{\cosh^2{x} - 1}} + C \\ &= -\frac{1}{1 + \sqrt{t^2 - 1}} + C\end{align*}
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    Re: Trig Substitution Integral

    Oh yeah. That was only slightly stupid... What I really meant to say was this:

    \int \cos^2 \theta d\theta = \frac{x}{2} + \frac{\sin2x}{4} + C = \frac{x + \sin x \cos x}{2} + C

    The right triangle part is where I am still struggling. I drew the triangle with theta as the angle, the hypotenuse labeled t and the x axis labeled a. Thus, \sec \theta = \frac{t}{a}

    When t = 2, \sec \theta = ??? This should be pretty obvious, but somehow I'm having a brain hemorrhage on this. Am I approaching this correctly? What am I missing?

    Thanks again.
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    Re: Trig Substitution Integral

    Quote Originally Posted by joatmon View Post
    Oh yeah. That was only slightly stupid... What I really meant to say was this:

    \int \cos^2 \theta d\theta = \frac{x}{2} + \frac{\sin2x}{4} + C = \frac{x + \sin x \cos x}{2} + C

    The right triangle part is where I am still struggling. I drew the triangle with theta as the angle, the hypotenuse labeled t and the x axis labeled a. Thus, \sec \theta = \frac{t}{a}

    When t = 2, \sec \theta = ??? This should be pretty obvious, but somehow I'm having a brain hemorrhage on this. Am I approaching this correctly? What am I missing?

    Thanks again.
    Good start on your triangle. I would let a = 1, since that's the substitution you used earlier. Use the Pythagorean Theorem to determine the vertical side of the triangle. What do you get?
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    Re: Trig Substitution Integral

    I think the values of \cos^{-1}{\left(\frac{1}{2}\right)} and  \cos^{-1}{\left(\frac{1}{\sqrt{2}\right)} are pretty standard.
    That being said, deriving them from a sketch is probably a good practice.
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