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Math Help - Integration with Trig. Sub.

  1. #1
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    Integration with Trig. Sub.

    The question-

    int [(-5x)/(x^2 + 5)^(3/2)] dx

    I have-

    x=sqrt(5)tan(theta)
    dx=sqrt(5)sec^2(theta)dtheta

    my solutions manual says that (x^2 + 5) = 5sec^2(theta) <---where does this come from. Is there some trig identity I'm not seeing?

    I've attached an image which may show the problem better.
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  2. #2
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    Quote Originally Posted by jlmills5 View Post
    The question-

    int [(-5x)/(x^2 + 5)^(3/2)] dx

    I have-

    x=sqrt(5)tan(theta)
    dx=sqrt(5)sec^2(theta)dtheta

    my solutions manual says that (x^2 + 5) = 5sec^2(theta) <---where does this come from. Is there some trig identity I'm not seeing?

    I've attached an image which may show the problem better.
    5\tan^2{t} + 5 = 5(\tan^2{t} + 1) = 5\sec^2{t}

    because of the pythagorean identity ...

    \tan^2{t} + 1 = \sec^2{t}
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  3. #3
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    Ah. I knew it would be something simple I wasn't seeing. Thank you!!!
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by jlmills5 View Post
    The question-

    int [(-5x)/(x^2 + 5)^(3/2)] dx

    I have-

    x=sqrt(5)tan(theta)
    dx=sqrt(5)sec^2(theta)dtheta

    my solutions manual says that (x^2 + 5) = 5sec^2(theta) <---where does this come from. Is there some trig identity I'm not seeing?

    I've attached an image which may show the problem better.
    -\int\frac{5x}{(x^2+5)^{3/2}}\,dx

    x=\sqrt{5}\tan(u) \implies dx=\sqrt{5}\sec^2(u)\,du

    Using the identity 1+\tan^2(\alpha)=\sec^2(\alpha),

    We have that (\sqrt{5}\tan(u))^2+5=5\tan^2(u)+5=5\sec^2(u).

    (5\sec^2(u))^{3/2}=5\sqrt{5}\sec^3(u)

    Putting it all together, we have:

    -\int\frac{5\sqrt{5}\tan(u)}{5\sqrt{5}\sec^3(u)}\cd  ot\sqrt{5}\sec^2(u)\,du = -\sqrt{5}\int\frac{\tan(u)}{\sec(u)}\,du=-\sqrt{5}\int\sin(u)\,du=\sqrt{5}\cos(u)

    Plug back in for u and you're done.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by jlmills5 View Post
    The question-

    int [(-5x)/(x^2 + 5)^(3/2)] dx

    I have-

    x=sqrt(5)tan(theta)
    dx=sqrt(5)sec^2(theta)dtheta

    my solutions manual says that (x^2 + 5) = 5sec^2(theta) <---where does this come from. Is there some trig identity I'm not seeing?

    I've attached an image which may show the problem better.
    However, it would be much simpler to do a regular u-sub (without using trig):

    Let u=x^2+5 \implies du=2x\,dx

    -\frac{5}{2}\int\frac{du}{u^{3/2}}=-\frac{5}{2}\cdot-2\cdot\frac{1}{u^{1/2}}=\frac{5}{\sqrt{u}}=\boxed{\frac{5}{\sqrt{x^2+5  }}}
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