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

WolframAlpha to the rescue. Click on Show Steps to see how they got it.

The WolframAlpha shows not a very elegant result for this example

$\displaystyle \displaystyle{\begin{aligned}\int\dfrac{du}{\cos^2 u+3}&=\int\dfrac{du}{\cos^2u+3(\sin^2u+\cos^2u)}=\ int\dfrac{du}{4\cos^2u+3\sin^2u}\\&=\dfrac{1}{4}\i nt\dfrac{1}{1+{\left(\dfrac{\sqrt3}{2}\tan{u}\righ t)\!}^2}\dfrac{du}{\cos^2u}=\left\{\begin{gathered }\dfrac{\sqrt3}{2}\tanu=t\hfill\\\dfrac{du}{\cos^2 u}=\dfrac{2t}{\sqrt3}\,dt\hfill\\\end{gathered}\ri ght\}\\&=\dfrac{1}{2\sqrt3}\int\dfrac{dt}{1+t^2}=\ dfrac{\sqrt3}{6}\arctan{t}+C\\&=\dfrac{\sqrt3}{6}\ arctan\!\left(\dfrac{\sqrt3}{2}\tan{u}\right)+C\en d{aligned}}$

Quote:

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Originally Posted by DeMath

The WolframAlpha shows not a very elegant result for this example

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

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

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

-Dan

Quote:

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Originally Posted by topsquark

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

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

-Dan

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No space between the \tan command and the u. Should have typed "\tan u" instead of "\tanu".

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....