# test & some help needed on LaTex.

• Nov 6th 2010, 05:02 AM
arccos
test & some help needed on LaTex.
$\displaystyle \displaystyle \oint^\frac{\pi}{4}_1 (\dfrac {\sqrt{(\sin (x +60^\circ)} + (\cos^2x)} {\csc{{3}{\pi}}x}) \,d\vec{x}$
Just a random string of math. How do i make the integration sign bigger,write $\displaystyle \pi$ 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.
$\displaystyle \begin{array}{rcl} \displaystyle \Huge\Sigma\oint^\frac{\pi}{4}_1 (\dfrac {\sqrt{(\sin (x +60^\circ)} + (\cos^2x)} {\csc(\frac{\pi}{2}\dfrac{22342}{{1}{\frac{2}{3}}} )x}) \,d\vec{x} =\\ \dfrac{tan^5x}{cosxsinx} + C \end{array}$
• Nov 6th 2010, 05:09 AM
Prove It
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 $\displaystyle \displaystyle \int_a^b{f(x)\,dx} = F(b) - F(a)$.

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 $\displaystyle \displaystyle \frac{\pi}{4}$.

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

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 $\displaystyle \displaystyle \sqrt{b^2 - 4ac}$.
• Nov 6th 2010, 05:09 AM
Plato
Quote:

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

Just a random string of math. How do i make the integration sign bigger,write $\displaystyle \pi$ over a number eg. pi/4 ,input the degree sign and extend the square root sign(if possible)?

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

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

Note the \displaystyle and the extra braces used in \sqrt{}
• Nov 6th 2010, 05:44 AM
Ackbeet
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

$\displaystyle \displaystyle \int^{\pi}_{1} \left(\dfrac {\sqrt{\sin (x +60^{3}) + \cos^{2}(x)}} {\csc(3x)}\right)dx$
• Nov 6th 2010, 05:51 AM
arccos
Quote:

Originally Posted by Plato
$$\displaystyle \int^\pi_1 (\dfrac {\sqrt{(\sin (x +60^3) + (\cos^2x)}} {\csc3x}) \,dx$$ gives

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

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...
• Nov 6th 2010, 05:55 AM
Unknown008
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}

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

Testing a little.

$\displaystyle a(b+c) + b(a + c) = ab + ac + ab + bc \\ = 2ab + ac + bc$

Nope, the \\ doesn't work outside the array command.
• Nov 6th 2010, 05:57 AM
Plato
Quote:

Originally Posted by arccos
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?
• Nov 6th 2010, 06:04 AM
arccos
Quote:

Originally Posted by Plato
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!
• Nov 6th 2010, 02:35 PM
Drexel28
Quote:

Originally Posted by Unknown008
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}

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

Testing a little.

$\displaystyle a(b+c) + b(a + c) = ab + ac + ab + bc \\ = 2ab + ac + bc$

Nope, the \\ doesn't work outside the array command.

I think it's just easier to do
\displaystyle \begin{aligned}x &= y\\ &= z\end{aligned}
• Nov 7th 2010, 05:47 AM
Unknown008
Hm... never saw this command. Thank you! (Smile)