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Math Help - Stuck on some integration

  1. #1
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    Question Stuck on some integration

    I'm currently stuck with trying to find the answer for the integral of 2cosec^2(2x) between the values of pi/4 and pi/6.

    I can integrate it and get -cot(2x) but when I put the values in I get a math error because cot(2x) is 1/tan(2x) and therefore if I put pi/4 in it will never have an answer.

    I can get the answer from the back of the book but I am more interested (and in need of knowing) in how to get it.
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  2. #2
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    Quote Originally Posted by Beard View Post
    I'm currently stuck with trying to find the answer for the integral of 2cosec^2(2x) between the values of pi/4 and pi/6.

    I can integrate it and get -cot(2x) but when I put the values in I get a math error because cot(2x) is 1/tan(2x) and therefore if I put pi/4 in it will never have an answer.

    I can get the answer from the back of the book but I am more interested (and in need of knowing) in how to get it.
    \cot\left(\frac{\pi}2\right) = \frac{\cos\left(\frac{\pi}2\right)}{\sin\left(\fra  c{\pi}2\right)} = 0.

    Just applying the definition of cot will tell you its 0. So you don't need the calculator
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    Quote Originally Posted by Isomorphism View Post
    \cot\left(\frac{\pi}2\right) = \frac{\cos\left(\frac{\pi}2\right)}{\sin\left(\fra  c{\pi}2\right)} = 0.

    Just applying the definition of cot will tell you its 0. So you don't need the calculator
    0 isn't the answer
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    Quote Originally Posted by Beard View Post
    0 isn't the answer
    I did not say 0 is the answer. But \cot\left(\frac{\pi}2\right) = 0

    \int_{\frac{\pi}6}^{\frac{\pi}4} 2\, \text{cosec}^2 2x \, dx = - \cot 2x \bigg{|}_{\frac{\pi}6}^{\frac{\pi}4} = - \cot\left(\frac{\pi}2\right) + \cot \left(\frac{\pi}3\right) = \frac1{\sqrt{3}}
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    Quote Originally Posted by Isomorphism View Post
    I did not say 0 is the answer. But \cot\left(\frac{\pi}2\right) = 0

    \int_{\frac{\pi}6}^{\frac{\pi}4} 2\, \text{cosec}^2 2x \, dx = - \cot 2x \bigg{|}_{\frac{\pi}6}^{\frac{\pi}4} = - \cot\left(\frac{\pi}2\right) + \cot \left(\frac{\pi}3\right) = \frac1{\sqrt{3}}
    Unfortunately wrong again and that is why I hate it.
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  6. #6
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    Quote Originally Posted by Beard View Post
    Unfortunately wrong again and that is why I hate it.
    I dont think it is wrong.On what basis do you say my answer is wrong?Is it because of the book? What does your book say? Probably they have given the answer in some other form.

    What do you expect the answer to be?
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    Quote Originally Posted by Isomorphism View Post
    I dont think it is wrong.On what basis do you say my answer is wrong?Is it because of the book? What does your book say? Probably they have given the answer in some other form.

    What do you expect the answer to be?
    The answer expects its to be (4 - pi)/8
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  8. #8
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    Quote Originally Posted by Beard View Post
    Unfortunately wrong again and that is why I hate it.
    It's correct.

    Did you try rationalising the denominator?

    \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}.
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  9. #9
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    I messed up

    Sorry, I was looking at the answer for the other question I was stuck on and both of you are correct
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