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    You have to use the substitution x = 2cosu

    I did that and you get this:


    but to change limits from x to u: u = arccos(x/2)

    0 -> pi/2
    1 -> pi/3

    integral of 1/sin^2(u) is -cotx, cot(x) is undefined for pi/2 ... did I do something wrong?
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    (Original post by Bealzibub)


    You have to use the substitution x = 2cosu

    I did that and you get this:


    but to change limits from x to u: u = arccos(x/2)

    0 -> pi/2
    1 -> pi/3

    integral of 1/sin^2(u) is -cotx, cot(x) is undefined for pi/2 ... did I do something wrong?
    Hey,

    When x = pi/2, cot(x) = 0. You can assume that. Alternatively, are you aware of hyperbolic functions by any chance?

    Edit:
    If you are not content of assuming it's true, then recall the following identity:
     \cot{(\frac{\pi}{2} - x}) = \tan({x}) if you put x = 0, then  \cot{(\frac{\pi}{2}}) = 0
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    (Original post by AMarques)
    Hey,

    When x = pi/2, cot(x) = 0. You can assume that. Alternatively, are you aware of hyperbolic functions by any chance?

    Edit:
    If you are not content of assuming it's true, then recall the following identity:
     \cot{(\frac{\pi}{2} - x}) = \tan({x}) if you put x = 0, then  \cot{(\frac{\pi}{2}}) = 0
    Yes, i'm aware of them. I solved it by substitution x = 2sinu then it also works just fine. It was an exam question so I wanted to know for sure why this particular substitution x=2cosu didn't work.

    Okay, thanks i'll assume cot(pi/2) = 0 from now on.
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    (Original post by Bealzibub)
    Yes, i'm aware of them. I solved it by substitution x = 2sinu then it also works just fine. It was an exam question so I wanted to know for sure why this particular substitution x=2cosu didn't work.

    Okay, thanks i'll assume cot(pi/2) = 0 from now on.
    Alright,

    For the hyperbolic approach, you can use the fact that  1 - \tanh^{2}{x} = sech^{2}{x} then just use a substitution of  x = 2\tanh{u}

    Personally, the answer came out nicer for me instead of the integral that you obtained, but it works both ways.
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    (Original post by AMarques)
    Alright,

    For the hyperbolic approach, you can use the fact that  1 - \tanh^{2}{x} = sech^{2}{x} then just use a substitution of  x = 2\tanh{u}

    Personally, the answer came out nicer for me instead of the integral that you obtained, but it works both ways.
    Bear in mind that the hyperbolic approach won't make any sense to someone doing straight maths because the hyperbolic functions aren't introduced until FM. The trig substitution works just fine
 
 
 
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