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Math Help - Stumped on trig integral

  1. #1
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    Stumped on trig integral

    Problem:
    \int \frac {1}{\cos u\cot u} du

    So I am kind of stumped with these Us in the equation. Our professor has been gracious enough to point us to an integral solving site, but I get different answers based on switching U for X.

    I don't think I need to switch any of the variables out, but I am trying to figure out where I want to go from here. It doesn't seem like I should use the substitution method on this, but I could be wrong. Could someone walk me through this?
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  2. #2
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    Quote Originally Posted by x0ne View Post
    Problem:
    \int \frac {1}{\cos u\cot u} du

    So I am kind of stumped with these Us in the equation. Our professor has been gracious enough to point us to an integral solving site, but I get different answers based on switching U for X.

    I don't think I need to switch any of the variables out, but I am trying to figure out where I want to go from here. It doesn't seem like I should use the substitution method on this, but I could be wrong. Could someone walk me through this?
    \int \frac{1}{\cos(u)\cot(u)}du=\int\frac{1}{\cos(u)\fr  ac{\cos(u)}{\sin(u)}}du=\int \frac{\sin(u)}{\cos^2(u)}du

    now this is a substitution let t=\cos(u) \implies dt=-\sin(u)du

    So we get

    \int \frac{-1}{t^2}dt=\int -t^{-2}dt=t^{-1}+c=\frac{1}{\cos(u)}+c=\sec(u)+c
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  3. #3
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    \frac{1}<br />
{{\cos (u)\cot (u)}} = \frac{{\sin (u)}}<br />
{{\cos ^2 (u)}}
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  4. #4
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    Quote Originally Posted by Plato View Post
    \frac{1}<br />
{{\cos (u)\cot (u)}} = \frac{{\sin (u)}}<br />
{{\cos ^2 (u)}}
    On this site (Wolfram Mathematica Online Integrator) it says the answer should be:
    x \sec u \tan u

    I know from the trig identities that sin/cos = tan and that would then leave 1/cos which is sec. My only gripe with their answer is that X that is upfront. Do you know how that is calculated?
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  5. #5
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    Quote Originally Posted by x0ne View Post
    On this site (Wolfram Mathematica Online Integrator) it says the answer should be:
    x \sec u \tan u

    I know from the trig identities that sin/cos = tan and that would then leave 1/cos which is sec. My only gripe with their answer is that X that is upfront. Do you know how that is calculated?

    No offense but I think you used it incorrectly.

    First where did the x come from? There are no x's in the original and you are not integrating with respect to x, but with respect to u.

    You can check that the answer I gave you is correct by taking its derviative.

    \frac{d}{du}\sec{u}=sec(u)\tan(u)=\frac{\sin(u)}{\  cos(u)\cos(u)}=\frac{1}{\cos(u)\cot(u)}

    The last step comes from multiplying the top and bottom of the fraction by

    \frac{1}{\sin(u)}
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