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Math Help - test & some help needed on LaTex.

  1. #1
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    test & some help needed on LaTex.

    \displaystyle \oint^\frac{\pi}{4}_1 (\dfrac {\sqrt{(\sin (x +60^\circ)} + (\cos^2x)} {\csc{{3}{\pi}}x}) \,d\vec{x}
    Just a random string of math. How do i make the integration sign bigger,write \pi over a number eg. pi/4 ,input the degree sign and extend the square root sign(if possible)?

    More random playing around, I wanted to see how complicated it can get.
    \begin{array}{rcl}<br /> <br />
\displaystyle \Huge\Sigma\oint^\frac{\pi}{4}_1 (\dfrac {\sqrt{(\sin (x +60^\circ)} + (\cos^2x)} {\csc(\frac{\pi}{2}\dfrac{22342}{{1}{\frac{2}{3}}}  )x}) \,d\vec{x} =\\ \dfrac{tan^5x}{cosxsinx} + C \end{array}
    Last edited by arccos; November 6th 2010 at 06:05 AM.
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  2. #2
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    To make everything bigger, use \displaystyle in your math tags before writing your code. E.g. \displaystyle \int_a^b{f(x)\,dx} = F(b) - F(a) gives \displaystyle \int_a^b{f(x)\,dx} = F(b) - F(a).

    To write a fraction, use \frac{}{}, with the numerator in the first set of curly brackets and the denominator in the second. E.g. \frac{\pi}{4} gives \displaystyle \frac{\pi}{4}.

    To input the degree sign, use ^{\circ}, e.g. 30^{\circ} gives \displaystyle 30^{\circ}.

    The square root sign is given by \sqrt{}, write everything you want under it inside the curly brackets. E.g. \sqrt{b^2 - 4ac} gives \displaystyle \sqrt{b^2 - 4ac}.
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  3. #3
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    Quote Originally Posted by arccos View Post
    \int^\pi_1 (\dfrac {\sqrt(\sin (x +60^3) + (\cos^2x)} {\csc3x}) \,dx

    Just a random string of math. How do i make the integration sign bigger,write \pi over a number eg. pi/4 ,input the degree sign and extend the square root sign(if possible)?
    [tex]\displaystyle \int^\pi_1 (\dfrac {\sqrt{(\sin (x +60^3) + (\cos^2x)}} {\csc3x}) \,dx [/tex] gives

     \displaystyle \int^\pi_1 (\dfrac {\sqrt{(\sin (x +60^3) + (\cos^2x)}} {\csc3x}) \,dx

    Note the \displaystyle and the extra braces used in \sqrt{}
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  4. #4
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    As a matter of style, I would probably write

    \displaystyle \int^{\pi}_{1} \left(\dfrac {\sqrt{\sin (x +60^{3}) + \cos^{2}(x)}} {\csc(3x)}\right)dx

    for

    \displaystyle \int^{\pi}_{1} \left(\dfrac {\sqrt{\sin (x +60^{3}) + \cos^{2}(x)}} {\csc(3x)}\right)dx<br />
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  5. #5
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    Quote Originally Posted by Plato View Post
    [tex]\displaystyle \int^\pi_1 (\dfrac {\sqrt{(\sin (x +60^3) + (\cos^2x)}} {\csc3x}) \,dx [/tex] gives

     \displaystyle \int^\pi_1 (\dfrac {\sqrt{(\sin (x +60^3) + (\cos^2x)}} {\csc3x}) \,dx

    Note the \displaystyle and the extra braces used in \sqrt{}
    Thanks for the help. Oh and how do i arrange my equations ? When I do a normal paragraphing by spacing it dosent work. There should be a code for this i guess...
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  6. #6
    MHF Contributor Unknown008's Avatar
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    You could end the LaTeX and start on another line with another LaTeX.

    However, you can use the table format too.

    \begin{array}{rcl}
    a(b+c) + b(a + c) &=& ab + ac + ab + bc \\
    &=& 2ab + ac + bc
    \end{array}

    \begin{array}{rcl}<br />
a(b+c) + b(a + c) &=& ab + ac + ab + bc \\<br />
&=& 2ab + ac + bc<br />
\end{array}

    Testing a little.

    a(b+c) + b(a + c) = ab + ac + ab + bc \\<br />
= 2ab + ac + bc

    Nope, the \\ doesn't work outside the array command.
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  7. #7
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    Quote Originally Posted by arccos View Post
    Thanks for the help. Oh and how do i arrange my equations ? When I do a normal paragraphing by spacing it dosent work. There should be a code for this i guess...
    I don't really understand what you are asking.
    Can you explain a bit more?
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  8. #8
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    Quote Originally Posted by Plato View Post
    I don't really understand what you are asking.
    Can you explain a bit more?
    Looks like Unknown008's input is what I wanted to know. I'm getting the hang of this!
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Unknown008 View Post
    You could end the LaTeX and start on another line with another LaTeX.

    However, you can use the table format too.

    \begin{array}{rcl}
    a(b+c) + b(a + c) &=& ab + ac + ab + bc \\
    &=& 2ab + ac + bc
    \end{array}

    \begin{array}{rcl}<br />
a(b+c) + b(a + c) &=& ab + ac + ab + bc \\<br />
&=& 2ab + ac + bc<br />
\end{array}

    Testing a little.

    a(b+c) + b(a + c) = ab + ac + ab + bc \\<br />
= 2ab + ac + bc

    Nope, the \\ doesn't work outside the array command.
    I think it's just easier to do
    \begin{aligned}x &= y\\ &= z\end{aligned}
    Last edited by Drexel28; November 14th 2010 at 09:45 AM.
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  10. #10
    MHF Contributor Unknown008's Avatar
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    Hm... never saw this command. Thank you!
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