I'm working through problems on my book and I have one problem, It seems like for the more complex functions they just memorize the table, which to me is pointless if you do not know how to get to the shortcut without using some integration techinque so I have a few problems that I would like a "hint" on what would be the first thing to do like integrations by part or substitution or the best way to expand or break down the integral so I can figure it out on my own

$\displaystyle \int{\frac{1}{3+y^2}}$

I seen one fellow expand the denom. out and have a square root come out I have no clue where that one came from.

$\displaystyle \int{\frac{1}{cos^3x}}$

No idea what to do here, I am not content with memorizing the shortcut or whatever it maybe be called in that table.

Those are all for now

2. Originally Posted by The Power
I'm working through problems on my book and I have one problem, It seems like for the more complex functions they just memorize the table, which to me is pointless if you do not know how to get to the shortcut without using some integration techinque so I have a few problems that I would like a "hint" on what would be the first thing to do like integrations by part or substitution or the best way to expand or break down the integral so I can figure it out on my own

$\displaystyle \int{\frac{1}{3+y^2}}$

I seen one fellow expand the denom. out and have a square root come out I have no clue where that one came from.

$\displaystyle \int{\frac{1}{cos^3x}}$

No idea what to do here, I am not content with memorizing the shortcut or whatever it maybe be called in that table.

Those are all for now
For 1

$\displaystyle \int {\frac{{dy}}{{3 + {y^2}}}} .$

$\displaystyle \frac{1} {{3 + {y^2}}} = \frac{1} {3} \cdot \frac{1} {{1 + \frac{{{y^2}}} {3}}} = \frac{1} {3} \cdot \frac{1} {{1 + {{\left( {\frac{y} {{\sqrt 3 }}} \right)}^2}}}.$

So $\displaystyle \int {\frac{{dy}}{{3 + {y^2}}}} = \frac{1}{3}\int {\frac{{dy}} {{1 + {{\left( {\frac{y}{{\sqrt 3 }}} \right)}^2}}}} = \left\{ \begin{gathered} \frac{y}{{\sqrt 3 }} = u, \hfill \\dy = \sqrt 3 du \hfill \\ \end{gathered} \right\} = \frac{{\sqrt 3 }}{3}\int {\frac{{du}}{{1 + {u^2}}}} =$

$\displaystyle = \frac{{\sqrt 3 }}{3}\arctan u + C = \frac{{\sqrt 3 }}{3}\arctan \frac{y}{{\sqrt 3 }} + C.$

3. I seen you try to explain this in a earlier thread but I do not see where the sqrt comes from

My problem is the reference for every problem is done by looking at the table of integrals my annoyance is how they get to these integrals in the table, or for test time sake's is it better just to accept this is how the table of integrals is and make reference and plug in for corresponding values?

I guess I just want to learn the steps leading to the final integrals in the table for the "non elementary functions"