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Another integral

  1. Jul 26, 2004 #1
    Hello again people, thank you for the help I recieved on the last integral I posted here but never got around to replying to, the help was much appreciated. However i've encountered another integral which for some reason I just cannot seem to solve, being:

    [itex]\int \frac{\cot x}{\sin x}\,dx[/itex]

    If anyone can aid me in solving this I would be very glad. Thanks in advance.
  2. jcsd
  3. Jul 26, 2004 #2
    This might help you to see it:

    {d \over {dx}}({1 \over {f(x)}}) = {{ - f'(x)} \over {f(x)^2 }}
  4. Jul 26, 2004 #3


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    In many situations, it's really simple just to express everything in terms of sine and cosine:

    [itex]\int \frac{\cos x}{\sin ^2 x}\,dx[/itex]

    Let [itex]u = \sin x[/itex], therefore [itex]dx = du/\cos x[/itex]. Making the substitution:

    [itex]\int \frac{du}{u^2} = -u^{-1} + C = -\csc x + C[/itex]
  5. Jul 26, 2004 #4
    Thank you very much, now you've just shown me I cannot believe how I missed that, however that tends to be how it always is. :wink:
  6. Jul 27, 2004 #5
    I always get frustrated when something was staring me in the face but I couldn't see it. That's why I thought a hint may be of more help that the 'full' solution.

    Paul. :wink:
  7. Jul 27, 2004 #6
    I'd always get frustrated when I couldn't see something staring me in the face. That's why I thought a hint may be of more help that the 'full' solution.

    Paul. :wink:
  8. Aug 7, 2004 #7
    or u can write the above integrl as

    INTEGRAL OF cosx cosec(sqr)x

    take cosec(sqr)x as 2nd function and integrate by parts ...gives u answer instantly...
  9. Aug 7, 2004 #8
    Some general advice: after you have an integral in a "standard looking form" (not even a standard form, for you wouldn't be reading my advice if it were in a standard form) try integration by parts keeping the ILATE rule in mind

    I = inverse trigonometric function
    L = logarithmic function
    A = algebraic function
    T = trigonometric function
    E = exponential function

    This order gives you an idea of which function to chose as u and which to chose v, when you wish to evaluate the integral [tex]\int u dv[/tex].

    \int udv = uv - \int vdu

    By the way, Dr. Brain are you from India? My guess is that you're in class 11/12. Correct me if I am wrong ;-)
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