Halloween and Calculus

Goistein

Junior Member
Joined
Oct 8, 2006
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I have only one day left to make my costume as Mr. Calculus. On the costume, I'm putting every derivative law and integral law I know. But I can't get the anti-derivative of x^x. Please help!
 
Goistein said:
I have only one day left to make my costume as Mr. Calculus. On the costume, I'm putting every derivative law and integral law I know. But I can't get the anti-derivative of x^x. Please help!
You say you're decorating with every integral you know. Since this is an integral that can't be done, then... can't you leave it off...?

Eliz.
 
OK, but I'm also looking for anti-derivatives of inverse trig functions. (All 6).
 
Denis said:
http://www.geocities.com/pkving4math2tor4/4_the_elem_transc_func/4_01_02_02_diffn_of_the_inv_trig_func.htm


Halloween favorite dessert: BOOberry pie and I SCREAM :roll:

All images on that page have invalid links - aka: broken
 
I think he was looking for "anti-derivatives" of the inverse-trig function.

That would be a good excercize for Mr. Calculus.

Hint:
Code:
[integral]sin^(-1)(x) dx

= x * sin[sup]-1[/sup]x - [integral][x/{sqrt(1-x^2})] dx

Now continue....
 
Any good calc texts has these in the cover.

\(\displaystyle \L\\\int{csc^{-1}(u)}du=ucsc^{-1}(u)du+ln|u+\sqrt{u^{2}-1}|+C\)

\(\displaystyle \L\\\int{sec^{-1}(u)}du=usec^{-1}(u)-ln|u+\sqrt{u^{2}-1}|+C\)

\(\displaystyle \L\\\int{cot^{-1}(u)}du=ucot^{-1}(u)+ln\sqrt{1+u^{2}}+C\)

\(\displaystyle \L\\\int{tan^{-1}(u)du}=utan^{-1}(u)-ln\sqrt{1+u^{2}}+C\)

\(\displaystyle \L\\\int{cos^{-1}(u)}du=ucos^{-1}(u)-\sqrt{1-u^{2}}+C\)

\(\displaystyle \L\\\int{sin^{-1}(u)}du=usin^{-1}(u)+\sqrt{1-u^{2}}+C\)
 
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