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Math Help - I need help integrating using trig substitutions

  1. #1
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    I need help integrating using trig substitutions

    The problem is to find the integration of sqrt(x^2 + 16) .
    I have been using x=4tan θ for the substitution and dx=4sec^2 θ .

    Please show your work.
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  2. #2
    MHF Contributor chisigma's Avatar
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    With the substitution \sqrt{x^{2} + 16} = x+z and a little computation [...] the integral...

    \displaystyle \int \sqrt{x^{2} + 16}\ dx (1)

    ... becomes the integral of a rational function in z...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by mlhuff12 View Post
    The problem is to find the integration of sqrt(x^2 + 16) .
    I have been using x=4tan θ for the substitution and dx=4sec^2 θ .

    Please show your work.
    If you draw a right-angled triangle, perpendicular sides...base=x, height=4

    then the hypotenuse length is \sqrt{x^2+4^2}

    \theta is the angle opposite the side of length 4.

    cos\theta=\frac{x}{\sqrt{x^2+16}}\ \Rightarrow\ \sqrt{x^2+16}=\frac{x}{cos\theta}

    tan\theta=\frac{4}{x}\ \Rightarrow\ x=\frac{4}{tan\theta}

    \frac{dx}{d\theta}=4\left(-cosec^2\theta\right)

    Therefore,

    \int_{\sqrt{x^2+16}}dx=4\int_{\frac{4}{tan\theta\ cos\theta}}\left(-cosec^2\theta\right)d\theta}

    etc
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  4. #4
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    Quote Originally Posted by mlhuff12 View Post
    The problem is to find the integration of sqrt(x^2 + 16) .
    I have been using x=4tan θ for the substitution and dx=4sec^2 θ .

    Please show your work.
    Hyperbolic substitution may be easier in this case...

    Let x = 4\sinh{t} so that dx = 4\cosh{t}\,dt.

    The integral \int{\sqrt{x^2 + 16}\,dx} becomes

    \int{\sqrt{(4\sinh{t})^2 + 16}\cdot 4\cosh{t}\,dt}

     = 4\int{\cosh{t}\sqrt{16\sinh^2{t} + 16}\,dt}

     = 4\int{\cosh{t}\sqrt{16(1 + \sinh^2{t})}\,dt}

     = 4\int{4\cosh{t}\sqrt{\cosh^2{t}}\,dt}

     = 16\int{\cosh^2{t}\,dt}

     = 16\int{\frac{1}{2} + \frac{\cosh{2t}}{2}\,dt}

     = 16\left(\frac{t}{2} + \frac{\sinh{2t}}{4}\right) + C

     = 8t + 4\sinh{2t} + C

     = 8t + 8\sinh{t}\cosh{t} + C

     = 8t + 8\sinh{t}\sqrt{1 + \sinh^2{t}} + C.


    Now remembering that x = 4\sinh{t} that means \sinh{t} = \frac{x}{4} and t = \sinh^{-1}\left(\frac{x}{4}\right) then

    8t + 8\sinh{t}\sqrt{1 + \sinh^2{t}} + C = 8\sinh^{-1}\left(\frac{x}{4}\right) + 8\left(\frac{x}{4}\right)\sqrt{1 + \left(\frac{x}{4}\right)^2} +C

     = 8\sinh^{-1}\left(\frac{x}{4}\right) + 2x\sqrt{1 + \frac{x^2}{16}} +C

     = 8\sinh^{-1}\left(\frac{x}{4}\right) + 2x\sqrt{\frac{16 + x^2}{16}} + C

     = 8\sinh^{-1}\left(\frac{x}{4}\right) + \frac{2x\sqrt{16 + x^2}}{4} + C

     = 8\sinh^{-1}\left(\frac{x}{4}\right) + \frac{x\sqrt{16 + x^2}}{2} + C

     = \frac{16\sinh^{-1}\left(\frac{x}{4}\right) + x\sqrt{16 + x^2}}{2} + C.
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  5. #5
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    I haven't learned anything about hyperbolic substitution yet. In my notes it says in the case of sqrt(x+a) to use atan θ as the substitution.
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  6. #6
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    You will get an equivalent answer (because you can convert \sinh^{-1}\left(\frac{x}{4}\right) to its logarithmic equivalent, but using trigonometric substitution in this case gets messy.
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  7. #7
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    Quote Originally Posted by mlhuff12 View Post
    I haven't learned anything about hyperbolic substitution yet. In my notes it says in the case of sqrt(x+a) to use atan θ as the substitution.
    if that is the case, then you should have worked your original substitution down to this point ...

    16 \int \sec^3{t} \, dt


    let's ignore the constant (16) for a bit and focus effort on the integral of secant cubed. using integration by parts ...

    u = \sec{t} ... dv = \sec^2{t} \, dt

    du = \sec{t} \tan{t} \, dt ... v = \tan{t}

    \int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec{t} \tan^2{t} \, dt<br />

    \int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec{t} (\sec^2{t} - 1) \, dt

    \int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec^3{t} - \sec{t} \, dt

    \int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec^3{t} \, dt + \int \sec{t} \, dt

    2\int \sec^3{t} \, dt = \sec{t} \tan{t} + \int \sec{t} \, dt

    \int \sec^3{t} \, dt = \frac{1}{2} \sec{t} \tan{t} + \frac{1}{2} \ln|\sec{t} + \tan{t}| + C

    put the constant (16) back ...

    16\int \sec^3{t} \, dt = 8\sec{t} \tan{t} + 8\ln|\sec{t} + \tan{t}| + C


    I'll leave you the task of back substituting.
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  8. #8
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    Thanx. So apparently I already had the right answer. However completely different than the back of my book says though.
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  9. #9
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    the "right" answer is in terms of x ... I told you that my work was incomplete.
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