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Math Help - Tough Antiderivative

  1. #1
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    Tough Antiderivative

    \int \frac{10x\cos x-10\sin x}{x^{2}}dx

    How would you go about finding this antiderivative?

    Thanks,
    Keith
    Last edited by Keithfert488; February 10th 2010 at 09:32 AM. Reason: added dx
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  2. #2
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    Quote Originally Posted by Keithfert488 View Post
    \int \frac{10x\cos x-10\sin x}{x^{2}}

    How would you go about finding this antiderivative?

    Thanks,
    Keith
    That is what I called a "tricky" question !
    Watch it carefully.
    Think about the quotient rule!
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  3. #3
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    So the quotient rule says {(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^{2}}

    So do I just integrate both sides of that with respect to x?

    Oh, I see now. So

    u=\sin x
    {u}'=\cos x

    and

    v=x
    {v}'=1


    So it'll end up being \int \frac{10x\cos x-10\sin x}{{x}^2}dx=\frac{10\sin x}{x}+C?
    Last edited by Keithfert488; February 10th 2010 at 09:32 AM. Reason: added dx and +C
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  4. #4
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    Quote Originally Posted by Keithfert488 View Post
    So the quotient rule says {(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^{2}}

    So do I just integrate both sides of that with respect to x?
    No.
    The function which you want to integrate it is just the derivative of a function.
    Find this function.
    Use the quoteint rule to find it.
    when you want to solve an integral of a function, and you know that derivative of this function, then the integral will be easy.
    \int f'(x) dx = \int \frac{d}{dx}\left(f(x)\right)=f(x)+C
    as an example:
    \int x dx = \int \frac{d}{dx} \left(\frac{1}{2}x^2\right)=\frac{1}{2}x^2+C.
    The function which you want to integrate, is just a derivative of another function.
    Try to find it.
    I hope you understood what I want to say.
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  5. #5
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    I understand. haha it just came to me. I get it. I edited in the answer.
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  6. #6
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    Quote Originally Posted by Keithfert488 View Post
    So the quotient rule says {(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^{2}}

    So do I just integrate both sides of that with respect to x?

    Oh, I see now. So

    u=\sin x
    {u}'=\cos x

    and

    v=x
    {v}'=1


    So it'll end up being \int \frac{10x\cos x-10\sin x}{{x}^2}=\frac{10\sin x}{x}?
    Correct.
    Do not forget +C and the dx.
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  7. #7
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    But also what I was trying to get at is trying to get an "inverse quotient rule" that can apply to the quotient rule much as integration by parts applies to the product rule.

    so

    {(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^2}

    \frac{u}{v}=\int\frac{du}{v}-\int\frac{udv}{v^2}

    Or would it be better just to try and recognize the quotient rule?


    Is there a rule like this? I have not learned it but I have learned substitution and integration by parts.
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  8. #8
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    Quote Originally Posted by Keithfert488 View Post
    But also what I was trying to get at is trying to get an "inverse quotient rule" that can apply to the quotient rule much as integration by parts applies to the product rule.

    so

    {(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^2}

    \frac{u}{v}=\int\frac{du}{v}-\int\frac{udv}{v^2}

    Or would it be better just to try and recognize the quotient rule?


    Is there a rule like this? I have not learned it but I have learned substitution and integration by parts.
    10\int \frac{cos(x)}{x} dx is scary.
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  9. #9
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    Quote Originally Posted by General View Post
    10\int \frac{cos(x)}{x} dx is scary.
    Is that even possible? Isn't it one of those that cannot be represented in terms of elementary functions?
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  10. #10
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    Quote Originally Posted by Keithfert488 View Post
    Is that even possible? Isn't it one of those that cannot be represented in terms of elementary functions?
    Yes. It is.
    See:
    integrate cos(x)/x - Wolfram|Alpha
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  11. #11
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    we just integrate by parts:

    \begin{aligned}<br />
   \int{\frac{x\cos x-\sin x}{x^{2}}\,dx}&=\int{\left( -\frac{1}{x} \right)'(x\cos x-\sin x)\,dx} \\ <br />
 & =\frac{\sin x-x\cos x}{x}+\cos (x)+k \\ <br />
 & =\frac{\sin x}{x}+k.<br />
\end{aligned}
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  12. #12
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    Quote Originally Posted by Krizalid View Post
    we just integrate by parts:

    \begin{aligned}<br />
   \int{\frac{x\cos x-\sin x}{x^{2}}\,dx}&=\int{\left( -\frac{1}{x} \right)'(x\cos x-\sin x)\,dx} \\ <br />
 & =\frac{\sin x-x\cos x}{x}+\cos (x)+k \\ <br />
 & =\frac{\sin x}{x}+k.<br />
\end{aligned}
    Wow. That is so much simpler than the way I did it. Thanks! I should have thought of that... =/
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  13. #13
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    there're lots of problems involving nasty integrands, but the great key here is to know when to use a good integration by parts.

    this is not the most important case, of course.
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