Alright guys, if someone could quickly show me how to do these two problems that'd be great. for some reason, I cannot figure them out!

The first one I can turn into sin(2x)csc(x) but then I get stuck - I'm not sure how to integrate that.

2. Remember that sin(2x) = 2sin(x)cos(x). Sub that in and solve.

Because you are supposted to work these yourself and sometimes have to explain them in front of the class, I give you hints instead (at first at least).

Hint for the 1st: How you can write sin(2x) with the help of x (not 2x, but x)? The triconometric formulae for 2x corners are something which are good to remember by heart.

EDIT: Yes, well looks like someone already told what sin(2x) is

EDIT: About 2: The integration is kind of correct if log is ln (natural logarithm). To me 'log' is base N, where N should be indicated by a subscript or it is defined other way in the text. Of course you might use a different kind of style there.

So what? You think he wouldn't have found it by himself?
http://en.wikipedia.org/wiki/Trig_identities
Learn as many of those as you can, they will be extremely useful in calculus courses.

As for #3, it's true that u = cos(x), but the rest is a very weird mess. Here's a jump-start for you. du/dx = -sin(x). You must solve for u and du, not dx, because you want to change the variable to u, not back to x. Then substitute u and du into the integral. What do you get?

Alright, just did problem #2 and that 2sin(x)cos(x) was what I needed. I wasn't thinking. I really do have to get these identities down...

The answer is 2sin(x) no?

Now to work on the third...

Yup 2sin(x).

With regards to the third problem, if you substitute cos(x) as u, then when everything cancels you end up with 1/u and the integral of that is log(u)... (we use log and ln interchangeably here). So then I substitute cos(x) for u again and I get -log|cos(x)| so I'm fairly certain that what I have all the way is right (I don't know if the + c is needed or not in this problem.)

Thing is, when you do the actual integration between the two points given (3c and sqrt(y)) I guess that I'm just going to have one bit LOONG answer with a lot of variables then? Something like this maybe:

-log|cos(3c)| + log|cos(sqrt(y))|

After integration I get ln|cos(x)|, it shouldn't be negative, and it should be ln not log as they are not the same thing. Also, the +c only exists after you integrate, but since you derive after integration, the +c goes away again. Anyway, the result is actually not so bad, since 3c is a constant and it's derivative is 0, you're just left with d/dy -ln|cos(sqrt(y))| which is simple!

Isn't it negative because the derivative of cos(x) is -sin(x)? So that carries through to the 1/u part...

I just checked it on a friend's TI-89 and his calculator gave me the exact same answer that I had up there...

Yea you're right, I forgot a negative sign. It's been a while since I last did calculus!

Lol. I was replying to you in engineering class, and someone looked over and said, "Whoa, she's hot! How did you get her to help you with math?!"

I was like, "Dude. You underestimate me..."

Thanks for your help tho dude.

Quote from Stang70Fastback :

Lol. I was replying to you in engineering class, and someone looked over and said, "Whoa, she's hot! How did you get her to help you with math?!"

LMAO!

Boy, it would have been great to have your own laptop when I was in school. Much easier than a piece of paper and a pencil