1. ## Help with Integration

Integrate:

2. Originally Posted by lacy1104
Integrate:
This does not have an antiderivative

3. This is what the problem really says. I thought by doing that integral was how you did it.

Show that for on [-1,1], the expression for aj is given by:

4. Originally Posted by lacy1104
This is what the problem really says. I thought by doing that integral was how you did it.

Show that for on [-1,1], the expression for aj is given by:
and what is a_j exactly?

5. Originally Posted by Jhevon
and what is a_j exactly?

6. Originally Posted by lacy1104
two more discrepancies here. in the first post you had $\sqrt{1 ~{\color{red}+}~ x^2}$ in the denominator, which made the integral impossible.

secondly, now this $T_i(x)$ appears and you never told us anything about it. Please try to describe the problem completely, we do not have your text book

7. Originally Posted by lacy1104
This is what the problem really says. I thought by doing that integral was how you did it.

Show that for on [-1,1], the expression for aj is given by:
I was wrong in the first entry. it is -

T0(x) = 1, T1(x)=x
In general, Ti(x)= cos[i arccos x], i > 0

It is posted on this website (its question 3b)
http://www.math.niu.edu/~dattab/math435/HW4.pdf

8. Originally Posted by lacy1104
I was wrong in the first entry. it is -

T0(x) = 1, T1(x)=x
In general, Ti(x)= cos[i arccos x], i > 0

It is posted on this website (its question 3b)
http://www.math.niu.edu/~dattab/math435/HW4.pdf
if it should have been minus, then the first integral is no problem. make a substitution of $u = \arccos x$, you will get an integral that you can do integration by parts on