1. ## Definite integral.

This one was a tad tricky for me, let me know if I am incorrect. I apologize in advance in that I do not know how to add numbers to the integral sign, couldn't find it on Latex.

Interval is [1/6] to [1/2]

$\displaystyle \int \csc{\pi}t \cot{\pi}t dt$

Let $\displaystyle u=\pi t$

$\displaystyle du=\pi dt$

$\displaystyle \frac{1}{\pi} \int \csc{(u)} \cot{(u)} du$

$\displaystyle \frac{1}{\pi} \int -\csc{(u)}$

$\displaystyle [ \frac{1}{\pi} \ast (-\csc{(\frac{1}{2}})) ] - [ \frac{1}{\pi} \ast (-\csc{(\frac{1}{6}})) ]$

Thanks for looking it over.

2. You have two options when doing a u substitution on a definite integral: you can either change the limits to the new variable, or, when you've computed the anti-derivative, change it back to the old variable and plug in the old limits. You made the mistake of plugging in the old limits with the u variable, which is like comparing apples to oranges. Do you follow me?

3. Oh man I just realized what I've done! I forgot that you're supposed to plug those values in where it is a variable, not where it is u! That was a bad slip. I'll redo it after I eat lunch.

By a mistake I treated u as the variable, when I was supposed to put pi t back in and THEN plug in the two numbers.

4. Just do this:

$\displaystyle u(1/6)=\pi/6,$ and

$\displaystyle u(1/2)=\pi/2.$ Plug those in thus:

$\displaystyle \dfrac{1}{\pi}\left[-\csc(\pi/2)-(-\csc(\pi/6))\right].$

5. *Edit- Sorry I was typing that all up as you replied*

$\displaystyle \int \csc{\pi}t \cot{\pi}t dt$

Let $\displaystyle u=\pi t$

$\displaystyle du=\pi dt$

$\displaystyle \frac{1}{\pi} \int \csc{(u)} \cot{(u)} du$

$\displaystyle \frac{1}{\pi} \int -\csc{(u)}$

$\displaystyle [\frac{1}{\pi} \ast (-\csc{(\pi\frac{1}{2}))} ] - [\frac{1}{\pi} \ast (-\csc{(\pi\frac{1}{6}))} ]$

Is this right, now?

6. Correct. Of course, you can simplify a bit...

7. Yeah I do things the longer way at first. Thanks so much for your help, Ackbeet. Glad I did it correctly.

8. You're welcome. Have a good one!

9. Originally Posted by Marconis
*Edit- Sorry I was typing that all up as you replied*

$\displaystyle \int \csc{\pi}t \cot{\pi}t dt$

Let $\displaystyle u=\pi t$

$\displaystyle du=\pi dt$

$\displaystyle \frac{1}{\pi} \int \csc{(u)} \cot{(u)} du$

$\displaystyle \frac{1}{\pi} \int -\csc{(u)}$
But you should not have the integral sign there. You have already done the integral. What you should have is either $\displaystyle \frac{1}{\pi} (-csc(u))+ C$
and then evaluate- the "C" will cancel

or, better, $\displaystyle \left[\frac{1}{\pi} (-csc(u)) \right]_{u= \pi/6}^{\pi/2}$

$\displaystyle [\frac{1}{\pi} \ast (-\csc{(\pi\frac{1}{2}))} ] - [\frac{1}{\pi} \ast (-\csc{(\pi\frac{1}{6}))} ]$

Is this right, now?

10. I need to break that habit. I always do that.

So guys...I'm trying to find the answer and my calculator keeps saying ERROR: DOMAIN. I've tried typing in Ackbeets way and my way, and it keeps happening.

11. Hmm. Let's clean up the notation a bit and see if that doesn't help out in terms of evaluating on the calculator:

$\displaystyle \displaystyle-\frac{1}{\pi} \csc\left(\frac{\pi}{2}\right) + \frac{1}{\pi} \csc\left(\frac{\pi}{6}\right)=\frac{1}{\pi}\left( \csc\left(\frac{\pi}{6}\right)-\csc\left(\frac{\pi}{2}\right)\right)=$

$\displaystyle \displaystyle\frac{1}{\pi}\left(\frac{1}{\sin(\pi/6)}-\frac{1}{\sin(\pi/2)}\right).$

Make sure your calculator is in radian mode, and try evaluating that. What kind of calculator do you have?

12. I'm getting .31831. Is that what you got?

I am using a TI-83. I prefer the TI-89 but our dept. doesn't let us use it.

13. Yep, that's what I get. And it's correct. See here.

So the "answer" posted in the OP is wrong.

14. THAT SITE IS AWESOME! Thanks! Now I can check all my answers on there.

15. Yep. WolframAlpha is a beaut. You're welcome. Have a good one!