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Math Help - Finding the indefinite integral

  1. #1
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    Finding the indefinite integral


    ⌡ cot^2(x) dx

    I approached this question by replacing cot^2 (x) w/ [cos^2(x)]/[sin^2(x)] => substituting sin^2(x) w/ u^2 => and ended up w/ -(1/3)cot(x) csc^2(x) + C

    The answer key just gives -cot (x) - x + C

    I have no idea of drawing any correlation so far except I didnt approach the trig sign from the correct position to begin with. Help with detailed algebraic work to this answer is what I ask please.
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  2. #2
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    Quote Originally Posted by ugkwan View Post

    ⌡ cot^2(x) dx

    I approached this question by replacing cot^2 (x) w/ [cos^2(x)]/[sin^2(x)] => substituting sin^2(x) w/ u^2 => and ended up w/ -(1/3)cot(x) csc^2(x) + C

    The answer key just gives -cot (x) - x + C

    I have no idea of drawing any correlation so far except I didnt approach the trig sign from the correct position to begin with. Help with detailed algebraic work to this answer is what I ask please.
    From the Pythagorean Identity: \cot^2 (x) = \text{cosec}^2 (x) - 1.

    And since the derivative of \cot (x) is - \text{cosec}^2 (x) ....
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  3. #3
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    Quote Originally Posted by ugkwan View Post

    ⌡ cot^2(x) dx

    I approached this question by replacing cot^2 (x) w/ [cos^2(x)]/[sin^2(x)] => substituting sin^2(x) w/ u^2 => and ended up w/ -(1/3)cot(x) csc^2(x) + C
    If you let u^2= sin^2(x), then u= sin(x) and du= cos(x) dx. Now you have two "cosines" in the problem so you will have \int\frac{cos(x)}{sin^2(x)} (cos(x)dx)= \int \frac{cos(x)}{u^2} du and you still have that "cos(x)" in the integral. You could then argue that cos(x)= \sqrt{1- sin^2(x)} and so change the integral to \int \frac{\sqrt{1- u^2}}{u^2} du but I don't that that's any easier!

    The answer key just gives -cot (x) - x + C

    I have no idea of drawing any correlation so far except I didnt approach the trig sign from the correct position to begin with. Help with detailed algebraic work to this answer is what I ask please.[/QUOTE]
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