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  1. #1
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    integration2

    integral((1-cos^4 x)/(cot x-tan x))

    in this i have taken 1 as (sin^2 x+cos^2 x)^2

    what should i do after that
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  2. #2
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    One way I guess would be to set t=tan x, so you get an integral of a rational expression. Then you would have to solve the resulting integral with partial fractions.
    By setting t=tan x I get

    \int \frac{1-\cos^4x}{\cot x - \tan x }\,dx =\int \frac{1-\frac{1}{(1+t^2)^2}}{\frac{1}{t}-t}\frac{\,dt}{1+t^2 }=\int \frac{t^4+2t^2}{\frac{(1-t^2)(1+t^2)^3}{t}}\,dt =\int \frac{t^3(t^2+2)}{(1-t^2)(1+t^2)^3}\,dt

    The rest of it is evaluating this integral. To begin with you should substitute for u=t^2. And so on. Hope this helps a little, and also hope someone has a more elegant solution.

    EDIT:

    Sorry, that way would have been hard, here is a better one. Note that:
    \frac{1}{\tan x}-\tan x = \frac{1-\tan^2 x}{\tan x}=\frac{2}{\tan 2x}
    \cos^2 x =\frac{1-\cos 2x}{2}
    So if you use that and substitute t=\cos 2x you should be able to solve the integral np.
    Last edited by Croat; June 5th 2011 at 02:33 AM.
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  3. #3
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    \displaystyle \begin{align*} \frac{1 - \cos^4{x}}{\cot{x} - \tan{x}} &= \frac{1 - \cos^4{x}}{\frac{\cos{x}}{\sin{x}} - \frac{\sin{x}}{\cos{x}}}\\ &= \frac{1 - \cos^4{x}}{\frac{\cos^2{x} - \sin^2{x}}{\sin{x}\cos{x}}} \\ &= \frac{(1 - \cos^2{x})(1 + \cos^2{x})}{\frac{\cos^2{x} - \sin^2{x}}{\sin{x}\cos{x}}} \\ &= \frac{\sin{x}\cos{x}(1 - \cos^2{x})(1 + \cos^2{x})}{\cos^2{x} - \sin^2{x}} \\ &= \frac{\sin^3{x}\cos{x}(1 + \cos^2{x})}{\cos^2{x} - \sin^2{x}} \\ &= \frac{\sin^3{x}\cos{x}(2 - \sin^2{x})}{1 - 2\sin^2{x}}\end{align*}

    So \displaystyle \int{\frac{1 - \cos^4{x}}{\cot{x} - \tan{x}}\,dx} = \int{\frac{\sin^3{x}(2 - \sin^2{x})\cos{x}}{1 - 2\sin^2{x}}\,dx}

    Make the substitution \displaystyle u = \sin{x} \implies du = \cos{x}\,dx and the integral becomes

    \displaystyle \int{\frac{u^3(2 - u^2)}{1 - 2u^2}\,du}

    Can you go from here?
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  4. #4
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    (nevermind, my arithmetic and latex don't mix).
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  5. #5
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    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \frac{1 - \cos^4{x}}{\cot{x} - \tan{x}} &= \frac{1 - \cos^4{x}}{\frac{\cos{x}}{\sin{x}} - \frac{\sin{x}}{\cos{x}}}\\ &= \frac{1 - \cos^4{x}}{\frac{\cos^2{x} - \sin^2{x}}{\sin{x}\cos{x}}} \\ &= \frac{(1 - \cos^2{x})(1 + \cos^2{x})}{\frac{\cos^2{x} - \sin^2{x}}{\sin{x}\cos{x}}} \\ &= \frac{\sin{x}\cos{x}(1 - \cos^2{x})(1 + \cos^2{x})}{\cos^2{x} - \sin^2{x}} \\ &= \frac{\sin^3{x}\cos{x}(1 + \cos^2{x})}{\cos^2{x} - \sin^2{x}} \\ &= \frac{\sin^3{x}\cos{x}(2 - \sin^2{x})}{1 - 2\sin^2{x}}\end{align*}

    So \displaystyle \int{\frac{1 - \cos^4{x}}{\cot{x} - \tan{x}}\,dx} = \int{\frac{\sin^3{x}(2 - \sin^2{x})\cos{x}}{1 - 2\sin^2{x}}\,dx}

    Make the substitution \displaystyle u = \sin{x} \implies du = \cos{x}\,dx and the integral becomes

    \displaystyle \int{\frac{u^3(2 - u^2)}{1 - 2u^2}\,du}

    Can you go from here?
    should i divide by u^2
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  6. #6
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    No, you should expand the numerator, then apply polynomial long division. It's probably easiest if you write it as \displaystyle \frac{u^5 - 2u^3}{2u^2 - 1}.
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    what should i do with the remainder
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  8. #8
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    Partial Fractions
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  9. #9
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    Hm, sorry for interrupting if you didn't like my solution, but since this is my first post I'd still like for it to be helpful.

    As mentioned, using the fact that:
    \tan 2x=\frac{2\tan x}{1-\tan^2 x}
    you can rewrite the denominator as:
    \cot x- \tan x=\frac{1-\tan^2 x}{\tan x}=2\frac{1-\tan^2x}{2\tan x}=\frac{2}{\tan 2x}.
    So with the other substitute for the cosine, you get:
    \int \frac{1-\cos^4x}{\cot x-\tan x}\,dx=\int \frac{1-(\frac{1+\cos 2x}{2})^2}{\frac{2}{\tan 2x}}
    Rewriting:
    \int \frac{1-\frac{1+2\cos 2x + \cos^2 2x}{4}}{2}\tan 2x \,dx=\frac{1}{8}\int (3-2\cos 2x-\cos^2 2x)\frac{\sin 2x}{\cos 2x}\,dx
    So the main idea is that the sine disappears with the substitution
    t=\cos 2x
    \,dt=-2\sin 2x\,dx
    \,dx=\frac{\,dt}{-2\sin 2x}
    So returning to the integral:
    \frac{1}{8}\int (3-2t-t^2)\frac{sin 2x}{t} \frac{\,dt}{-2\sin2x}=
    =\frac{1}{16}\int \frac{t^2+2t-3}{t}\,dt
    Which should prove a fairly easy integral.

    Well, anyway... if you're going the other way still consider substituting at least t=u^2. Good luck.
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