Question

I'm having trouble with this definite integral.
a=1/48
b=1/16

csc(8πt) cot(8πt) dt

Answer 1

\[\int\limits_{1/48}^{1/16}\csc(8πt)*\cot(8πt)*dt\]

Answer 2

anybody......

Answer 3

I did every one of them from my homework without any trouble, but can't solve this one...

So if I make u=8*pi*t
du=8pi*dt
(du/8pi)=dt

so:
\[\int\limits_{1/48}^{1/16} \csc(u)*\cot(u)*(du/8π)\]

Answer 4

Hint, the derivative of csc(u) = -cot(u)csc(u)

Answer 5

Aren't I supposed to take the antiderivative?

Answer 6

Oh wait, I think i see.... Gimme a sec, don't go anywhere please!

Answer 7

You're on the right path, but remember to change your limits!

u = 8 pi t
so u(1/16) = 8 pi (1/16) = pi/2
and
u(1/48) = 8 pi (1/48) = ... etc.

THEN you can go and do the rest of the integral, which should be easy from here on in. :)

Answer 8

Yeah I see that^^, Thanks, but I can't just plug those numbers into csc(u)*cot(u)*(du/8pi), right?

I have to plug those upper and lower limits into the antiderivative of csc(u)*cot(u)*(du/8pi), which is what i am struggling with.

Answer 9

Oh my gosh, I'm such an idiot. i see. Now your hint makes sense to me!

Answer 10

f(x)= -csc(x)+(x/8π) + C
f'(x)= csc(x) * cot(x) * (dx/8pi)


If my calculations are correct.

Answer 11

Not quite...

if f(x) = csc(u)
then f'(x) = -cot(u)*csc(u)

Answer 12

And you already reduced your integral to

integral of 1/(8pi) cot(u)*csc(u) du

Answer 13

So you're like... right on the doorstep of this.

Answer 14

Now i'm just going to plug in u(1/46) and u(1/16) into -csc(u)+((u/8pi)???

Answer 15

So now do I plug in the upper and lower limits (u(1/16) and u(1/46)) into
-csc(u)+((u/8pi)
and find the difference????

Answer 16

I can't figure it out! damnit

Answer 17

[-csc(pi/6)+((pi/6)/(8pi)) - -csc(pi/2)+((pi/2)/(8pi))]

= [-15/16 - -95/48]
= 25/24


Which IS NOT RIGHT.........

Okay, F*ck this, I've spent two hours on this and i have other sh*t to do...