1. ## Toughie Integral.

$\displaystyle I(x):=\frac{1}{6}\int_{2x^3}^{3x^2} \frac{sin(t)}{t}dt$ for $\displaystyle 0<x<\frac{3}{2}.$
I know that $\displaystyle Si(x)=\int_0^{x}\frac{sin(t)}{t}dt$. I can't see how this helps!!

Integration by parts isn't useful either as it makes an infinitely long sequence.

Macluarin expansions also don't help. They make something very messy.

gah!!

2. No elementary primitive.

3. What do you mean by that?

Is there a special trick to integrate it?

It seems like there is something to it because it has it's own definition for it's integral: Si(x).

4. No trick, you can't find a primitive.

Perhaps this is a question involving the FTC.

5. What does FTC stand for?

I tried googling it but it didn't work =S

Is a primitive the same as an antiderivative?

And, FTC means Fundamental Theorem of Calculus; but dunno what you want to do with your problem.

7. umm, I want to find an expression for the integral.

I have the answer, but I can't work out how to get there.

I get stuck on having $\displaystyle sin(3x^2)$ and $\displaystyle sin(2x^3)$. Is there a way to change these into just $\displaystyle sin(x)$? (my question gives the answer in terms of just $\displaystyle sin(x)$).

Double angle formulae don't work for this part either...

8. Please, give me the answer, I'm pretty sure what is this.

9. Okay then, here's the rest of the question:

Show that $\displaystyle \frac{dI}{dx}=(1-x)sin(x)$ and deduce that $\displaystyle I(x)=1-cos(x)+xcos(x)-sin(x)$
.

10. Originally Posted by Showcase_22
Okay then, here's the rest of the question:

.
So differentiate $\displaystyle I(x)=\frac 16 \int_{2x^3}^{3x^2} \frac{\sin(t)}{t} ~dt$

Let $\displaystyle F(t)$ be an antiderivative of $\displaystyle \frac{\sin(t)}{t}$

Then $\displaystyle I(x)=\frac 16 (F(3x^2)-F(2x^3))$

Now you know that $\displaystyle F'(t)=\frac{\sin(t)}{t}$ by definition of an antiderivative.

So $\displaystyle \frac{dI}{dx}=x F'(3x^2)-x^2 F'(2x^3)$ by the chain rule of differentiation.

$\displaystyle \frac{dI}{dx}=x \cdot \frac{\sin(3x^2)}{3x^2}-x^2 \cdot \frac{\sin(2x^3)}{2x^3}=\frac{1}{x} \cdot \sin(3x^2)-\frac 1x \cdot \sin(2x^3)$

I'm sorry I have to go to eat... I'll think about the rest later

11. Well there's no need.

I walked into todays lecture and he changed the question completely!

I like your method Moo, i'll remember that for another time (hopefully for a similar question in the future!).

If anyone can actually integrate this between the limits I would really like to see how it's done. It's really quite puzzling.