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Math Help - Integral of cos(x)/sin^2(x).

  1. #1
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    Integral of cos(x)/sin^2(x).

    Hello guys here is my problem uploaded: , see the attached image of my scanned paper.Integral of cos(x)/sin^2(x).-img-001.jpg
    Last edited by mr fantastic; January 15th 2011 at 12:52 AM. Reason: Title.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Riazy View Post
    Click image for larger version. 

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ID:	20444Hello guys here is my problem uploaded:
    Your second attempt is almost correct (you're missing a minus sign on the final answer...and its csc x, not cot x).

    There are two ways to go about this:

    1) Make the substitution u=\sin x. Then \displaystyle\int\dfrac{\cos x}{\sin^2x}\,dx\xrightarrow{u=\sin x}{}\int\dfrac{\,du}{u^2}=\ldots

    2) Observe that \dfrac{\cos x}{\sin^2 x}=\dfrac{1}{\sin x}\cdot\dfrac{\cos x}{\sin x}=\csc x\cot x and recall that \dfrac{\,d}{\,dx}\left[\csc x\right]=-\csc x\cot x.

    So either way you do this, you should end up with \displaystyle\int\dfrac{\cos x}{\sin^2x}\,dx=-\csc x+C.
    Last edited by Chris L T521; January 15th 2011 at 12:07 AM. Reason: fixed a couple errors.
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  3. #3
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    Your substitution should be \displaystyle u = \sin{x}, not \displaystyle u = \sin^2{x}. Then the integral reduces to

    \displaystyle \int{u^{-2}\,du}.
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    I have a problem though, My key answer book gives the answer -1/sinx + c. I am a bit confused now, are there different answers?
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    You should know that \displaystyle \frac{1}{\sin{x}} = \csc{x}...
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    Oh, over here in sweden we don't use csc x, not atleast in my bachelor civil engineering program, Thanks alot BTW! both of u
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  7. #7
    Super Member Quacky's Avatar
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    I would have done this slightly differently.

    It has been proven that for any function f(x), \displaystyle \int f'(x)f(x)^n dx=\frac{1}{n+1}[f(x)^{n+1}]+c

    This applies here if you write this as:

    [Cos(x)][Sin(x)^{-2}]

    as f(x) = sin x; n = -2; f'(x) = Cos(x)

    Substituting into:

    \frac{1}{n+1}[f(x)^{n+1}]+c

    Gives \frac{1}{-2+1}[Sin(x)^{-2+1}]+c

    =\frac{1}{-1}[Sin(x)^{-1}]+c

    =\displaystyle\frac{-1}{Sin(x)}

    =-\csc(x) + c
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    Hello, Riazy!


    I did it like this . . .

    \displaystyle \int \frac{\cos x}{\sin^2\!x}\,dx \;=\;\int\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}\,dx \;=\;\int \csc x \cot x\,dx \;=\;-\csc x + C

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    Quote Originally Posted by Soroban View Post
    Hello, Riazy!


    I did it like this . . .

    \displaystyle \int \frac{\cos x}{\sin^2\!x}\,dx \;=\;\int\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}\,dx \;=\;\int \csc x \cot x\,dx \;=\;-\csc x + C

    I don't know how you can all remember integrals like \displaystyle \int{\csc{x}\cot{x}\,dx} or \displaystyle \int{f'(x)[f(x)]^n\,dx} = \frac{1}{n+1}[f(x)]^{n+1} + C off the top of your head. I prefer to remember the procedures, and in this case, to reduce everything to sines and cosines and use substitution - at least then I only have to remember a couple of fundamental integrals
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    Super Member Quacky's Avatar
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    Quote Originally Posted by Prove It View Post
    I don't know how you can all remember integrals like \displaystyle \int{\csc{x}\cot{x}\,dx} or \displaystyle \int{f'(x)\,f^n(x)\,dx} = \frac{1}{n+1}f^{n+1}(x) + C off the top of your head. I prefer to remember the procedures, and in this case, to reduce everything to sines and cosines and use substitution - at least then I only have to remember a couple of integrals
    I don't always remember the integrals, I often just remember how to derive them from differentiation.
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  11. #11
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    You should be able to remember"
    If, in the integral \int sin^n(x)cos^m(x)dx, n is odd, say n= 2k+ 1, you can factor out one "sine" to use with the differential: \int sin^{2k}(x) cos^m(x) (sin(x)dx), then use sin^2(x)= 1- cos^2(x) to get
    \int (1- cos^2(x))^k cos^m(x)(sin(x)dx)
    and now the substitution u= cos(x), so that du= -sin(x)dx, makes that a polynomial integtration:
    -\int (1- u^2)^k u^m u
    Of course, if m is odd, you can do the same thing, switching "sine" and "cosine".

    This also works if either m or n (or both) are negative but gives a "rational function" integtral rather than a polynomial integral.

    The specific formulas others gave may be quicker for specific values of m or n, but that is, in my opinion, the simplest general method.


    (If both m and n are even, you can use sin^2(x)= (1/2)(1- cos(2x) and/or cos^2(x)= (1/2)(1+ cos(2x)) to reduce the powers.)
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  12. #12
    Super Member Quacky's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    You should be able to remember"

    Of course, if m is odd, you can do the same thing, switching "sine" and "cosine".

    This also works if either m or n (or both) are negative but gives a "rational function" integtral rather than a polynomial integral.

    The specific formulas others gave may be quicker for specific values of m or n, but that is, in my opinion, the simplest general method.


    (If both m and n are even, you can use sin^2(x)= (1/2)(1- cos(2x) and/or cos^2(x)= (1/2)(1+ cos(2x)) to reduce the powers.)
    Oh! Is that ALL? Well, thanks for clearing that up! That's so much simpler and easier to remember!
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    Quote Originally Posted by HallsofIvy View Post
    You should be able to remember"

    Of course, if m is odd, you can do the same thing, switching "sine" and "cosine".

    This also works if either m or n (or both) are negative but gives a "rational function" integtral rather than a polynomial integral.

    The specific formulas others gave may be quicker for specific values of m or n, but that is, in my opinion, the simplest general method.


    (If both m and n are even, you can use sin^2(x)= (1/2)(1- cos(2x) and/or cos^2(x)= (1/2)(1+ cos(2x)) to reduce the powers.)
    I don't know HOW I could have been so blind... How DID I get by just remembering the integrals of \displaystyle \sin{x}, \cos{x} and \displaystyle \sec^2{x}? :P
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    Quote Originally Posted by Prove It View Post
    I don't know how you can all remember integrals like \int{f'(x)[f(x)]^n\,dx} = \frac{1}{n+1}[f(x)]^{n+1} + C
    It's very hard to forget that integral once you've read it 'backwards' (well, sort of):

    \left(\frac{1}{n+1}[f(x)]^{n+1}\right)' = f'(x)[f(x)]^n

    ... because it's very a chain rule picturesque. Or I memorise things in weird ways!
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  15. #15
    Super Member Quacky's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    It's very hard to forget that integral once you've read it 'backwards' (well, sort of):

    \left(\frac{1}{n+1}[f(x)]^{n+1}\right)' = f'(x)[f(x)]^n

    ... because it's very a chain rule picturesque. Or I memorise things in weird ways!
    I do something similar to learn it, so you're only about as crazy as I am. Which is probably not very reassuring.
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