# integrate the function....

• Feb 14th 2011, 10:48 AM
slapmaxwell1
integrate the function....
ok i have a problem and i dont remember how to integrate...

1/(cos^2(u) + 3)du

the function should read 1 over the function of cosine squared of u +3
• Feb 14th 2011, 11:01 AM
Ackbeet
WolframAlpha to the rescue. Click on Show Steps to see how they got it.
• Feb 14th 2011, 11:42 AM
DeMath
The WolframAlpha shows not a very elegant result for this example

\displaystyle{\begin{aligned}\int\dfrac{du}{\cos^2 u+3}&=\int\dfrac{du}{\cos^2u+3(\sin^2u+\cos^2u)}=\ int\dfrac{du}{4\cos^2u+3\sin^2u}\\&=\dfrac{1}{4}\i nt\dfrac{1}{1+{\left(\dfrac{\sqrt3}{2}\tan{u}\righ t)\!}^2}\dfrac{du}{\cos^2u}=\left\{\begin{gathered }\dfrac{\sqrt3}{2}\tanu=t\hfill\\\dfrac{du}{\cos^2 u}=\dfrac{2t}{\sqrt3}\,dt\hfill\\\end{gathered}\ri ght\}\\&=\dfrac{1}{2\sqrt3}\int\dfrac{dt}{1+t^2}=\ dfrac{\sqrt3}{6}\arctan{t}+C\\&=\dfrac{\sqrt3}{6}\ arctan\!\left(\dfrac{\sqrt3}{2}\tan{u}\right)+C\en d{aligned}}
• Feb 14th 2011, 12:43 PM
topsquark
Quote:

Originally Posted by DeMath
The WolframAlpha shows not a very elegant result for this example

$\displaystyle\left\{\begin{gathered}\dfrac{\sqrt3} {2}\tanu=t\hfill\\\dfrac{du}{\cos^2u}=\dfrac{2t}{\ sqrt3}\,dt\hfill\\\end{gathered}\right\}$

You meant
$\displaystyle t = \frac{\sqrt{3}}{2} tan (u)$

I don't know why that didn't come out.

-Dan
• Feb 14th 2011, 12:52 PM
Ackbeet
Quote:

Originally Posted by topsquark
You meant
$\displaystyle t = \frac{\sqrt{3}}{2} tan (u)$

I don't know why that didn't come out.

-Dan

No space between the \tan command and the u. Should have typed "\tan u" instead of "\tanu".
• Feb 14th 2011, 03:14 PM
slapmaxwell1
ok so i need to get a pocket version of wolfram..lol thanks everybody i thought i was losing it..when i couldnt figure it out....