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\[f(x,y)= \int\limits_{y}^{x} \cos (t^9) dt\]

this is what i got

\[1/x^9 \sin(x^9)+1/y^9 \sin(y^9) +C\]

not sure if its right though

ohhh..

i am suppose to integrate by parts right?

1/9sinx^9-1/9sinx^9+c

i totally spaced out on that

1/9sinx^9-1/9siny^9+c

yes you are right no need +c

so we don't integrate this by u substitution?

or some other integration method?

no need substitution

hmm.......... actually

so what does that mean?

http://www.wolframalpha.com/input/?i=integral%28cos%28t^9%29dt%29

yeah we haven't learned this and there isn't an example in a book

http://how-to-spell-ridiculous.com/

You are going to need an incomplete gamma or knowledge of the Heaviside to solve this.

hahahaha thanks ScienceWiz

<3

yes

where F(x)= integral of x

Thoughts?

yeah I am kind of confused because I am a bit rusty on the FTC

FTC just states The derivative of the Integral is your original equation..

I need to find the first partial derivatives fx and fy

so f(x) = cos (x ^9) and -f(y)= -cos(y^9)? because of FTC

Pretty much.

that seems much easier :) I appreciate your help!!
THANKS