Question

Please! Help!(:
Find the antiderivative by integrating the function
f(x)=csc(4x)cot(4x)

Answer 1

\[\large \int\limits \csc(4x)\cot(4x)dx\]

Hmmm, let's try a `U substitution` and see what happens.
If we let \(\large u=\csc(4x)\), I think it will lead somewhere nice.

We could try letting the other term be our u, just for the sake of trial and error, since that is the process you would probably take on this :) It would end up being the wrong sub, but it's the process that matters.

Taking the derivative of our substitution gives us,\[\large du=-4\csc(4x)\cot(4x)dx\]
If you're not familiar with that derivative, then the last step maybe have been a little confusing. Understand what I've done so far? :o

Answer 2

kind of? where did that -4 come from?

Answer 3

\[\int\limits_{}^{} \csc(4\theta)\cot(4\theta) d \theta \]
That was the real function

Answer 4

\[\large \left(\csc x\right)' \qquad=\qquad -\csc x \cot x\]
The 4 came from the chain rule. :D

Answer 5

I have to evaluate the integral.

Answer 6

Yes, that's what we're working on.
I just used x's instead of theta's.

Answer 7

I guess a substitution like that is really not necessary..
Because if you know this derivative,\[\large \left(\csc x\right)' \qquad=\qquad -\csc x \cot x\]

Then certainly you know this integral,\[\large \int\limits -\csc x \cot x dx\]

Ok my mistake, all we really need to do is deal with the 4theta inside.

Answer 8

oh okay. so howd you get from
\[\csc(4x)\cot(4x)\]
to
\[(cscx)'=-\csc(x) \cot(x)\]

Answer 9

I'm so confused. :( haha

Answer 10

This is just a derivative rule that I was mentioning,\[(cscx)'=-\csc(x) \cot(x)\]

If you haven't learned that yet, then maybe we should go about doing this the more tedious way ~ Converting cosecant and cotangent to sines and cosines and then do a u-substitution. :D

Answer 11

could we convert then try the u-sub.?

Answer 12

\[\large \int\limits \csc(4\theta)\cot(4\theta) d \theta \qquad=\qquad \int\limits \frac{1}{\sin(4\theta)}\left(\frac{\cos(4\theta)}{\sin(4\theta)}\right)d \theta\]

So converting to sines and cosines gives us,\[\large \int\limits \frac{\cos(4\theta)d \theta}{\sin^2(4\theta)}\]Following along so far? :)

Answer 13

yes(:

Answer 14

Letting,\[\large u=\sin(4\theta)\]
Taking the derivative of our substitution gives us,\[\large du=4\cos(4\theta)d \theta\]

Answer 15

So with our last result right there, we want to divide each side by 4,\[\large \frac{1}{4}du=\cos(4\theta)d \theta\]

Answer 16

okay.

Answer 17

where do we go from there?

Answer 18

\[\large \color{purple}{u=\sin(4\theta)}\]\[\large \color{green}{\frac{1}{4}du=\cos(4\theta)d \theta}\]

We'll use this information to rewrite our integral in terms of u.\[\large \int\limits\limits \frac{\color{green}{\cos(4\theta)d \theta}}{\color{purple}{\sin(4\theta)}^2} \qquad\to\qquad \int\limits\limits \frac{\color{green}{\frac{1}{4}du}}{\color{purple}{u}^2} \]

Answer 19

woah. okay haha.

Answer 20

We'll rewrite our u using rules of exponents,
\[\large \frac{1}{4}\int\limits u^{-2}du\]
And from here we can apply the Power Rule for Integration to do the integration portion.

Answer 21

What anti-derivative does that produce? :)

Answer 22

-2u ^-2+1?

Answer 23

Umm I'm not sure where the -2 is coming from, in front.

Answer 24

the power?

Answer 25

Remember, with integrals, you divide by the NEW EXPONENT.
Not the -2, but the -1 that we end up with.

Answer 26

2u^-1

Answer 27

Ignoring the 1/4 fraction for a sec, here is what we want to be doing,
\[\large \int\limits u^{-2} \qquad=\qquad \frac{u^{-2+1}}{-2+1} \qquad=\qquad -u^{-1}\]

Answer 28

ohhhh. haha.

Answer 29

okay I see where I went wrong. I was applying the power rule as a derivative. lol

Answer 30

Ah :)

Answer 31

So it looks like we end up with,\[\large \frac{1}{4}\int\limits\limits u^{-2}du \qquad=\qquad -\frac{1}{4}u^{-1}\]We'll want to do a little work to undo our substitution, and then write it in terms of csc or cot if we're able to.

Answer 32

so. \[-\frac{ 1 }{ 4 }\sin(4\theta)^{-1}\]

Answer 33

\[\large -\frac{1}{4}\left(\frac{1}{\sin(4\theta)}\right)+C\]Yup looks good!
Remember what 1/sin gives us?

Answer 34

csc?

Answer 35

\[\large -\frac{1}{4}\csc(4\theta)+C\]

Yay good job! \c:/

Answer 36

Omg that was so hard! lol thank you so much!!!(:

Answer 37

np

Answer 38

do you know if you could help me with a similar prob/? @zepdrix

Answer 39

Sure, post it in a new thread.
This one is too long :U