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Math Help - Integrals

  1. #1
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    Integrals

    Integral of (cosx) / (1+sin^2 x + sin^4 x )

    is there a way to do this without the use of a calculator? thanks
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  2. #2
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    Yes, but it's a loo-loo. You can let u=sin(x), \;\ du=cos(x)dx

    \int\frac{1}{1+u^{2}+u^{4}}du

    \int\frac{1}{(u^{2}+u+1)(u^{2}-u+1)}du

    You can use partial fractions. One way. Kriz will probably be along with some show off method
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  3. #3
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    Well sometimes we can turn nasty things into beauty ones, but this one is very nasty.

    I haven't developed a quickly way to get this faster, but it looks like the straightforward way can tackle it. (Not nicely, of course.)
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  4. #4
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    Weierstrass trick also does not work here well. It leads to a messy integral too.
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  5. #5
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    Hello, xfyz!

    I had a wild idea . . . but it got me nowhere.
    . .
    (But this trick has worked for me before.)

    \int \frac{\cos x\,dx}{1 + \sin^2\!x + \sin^4\!x}

    Multiply by \frac{\cos^2\!x}{1-\sin^2\!x} . . . . Note that it equals 1.

    . . \frac{\cos x}{1 + \sin^2\!x + \sin^4\!x}\cdot\frac{\cos^2\!x}{1-\sin^2\!x} \;=\;\frac{\cos^3\!x}{1 - \sin^6\!x} . . . . very cute. . But now what?

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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, xfyz!

    I had a wild idea . . . but it got me nowhere.
    . .
    (But this trick has worked for me before.)


    Multiply by \frac{\cos^2\!x}{1-\sin^2\!x} . . . . Note that it equals 1.

    . . \frac{\cos x}{1 + \sin^2\!x + \sin^4\!x}\cdot\frac{\cos^2\!x}{1-\sin^2\!x} \;=\;\frac{\cos^3\!x}{1 - \sin^6\!x} .

    nice. but i don't really see how to take it from here either. i was going to tell you to expand the denominator using the difference of two cubes formula, but then i realized that would just give you back the original function.
    . . . very cute. . But now what?
    hehe, that made me laugh
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  7. #7
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    \frac{\cos^3 x}{1-\sin^6 x} = \frac{(1-\sin^2 x)\cos x}{1-\sin^6 x}
    Let t = \sin x then,
    \int \frac{1-t^2}{1-t^6} dt
    But it is still ugly.
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  8. #8
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    Even doing Soroban's suggestion does not look nice.

    The "best" way was provided by galactus, 'cause you'll get known integrals and calculations won't get nasty.
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