Question

Show that the two formulas are equivalent integral(cscxdx = -ln|cscx + cotx| + C)and integral(cscxdx = ln|cscx - cotx| + C )

Answer 1

This one is a little tricky. We'll have to solve this integral using 2 separate U substitutions and we'll show that we get 2 different answers for the integral (the one's that they're asking for).

Answer 2

\[\large \int\limits \csc x dx\]We'll start by multiplying the top and bottom by a weird looking ..thing. Keep in mind this won't change the value of the integral since it's techincally just "1".\[\large \int\limits \csc x dx \left(\frac{\csc x+\cot x}{\csc x + \cot x}\right) \quad = \quad \int\limits \frac{\csc^2x+\csc x \cot x}{\csc x + \cot x}dx\]

Answer 3

From here, we'll apply a U-substitution, letting our denominator be our U.\[\large \color{orangered}{u=\csc x + \cot x}\]To get your du, you might have to look these up to verify (since they don't come up as often).\[\large du=-\csc x \cot x - \csc^2 x dx\]\[\large \color{orangered}{-du=\csc^2x + \csc x \cot x dx}\]

Answer 4

Plugging in the orange pieces gives us,\[\large \int\limits \frac{-du}{u}\]

Answer 5

Integrating with respect to u gives us,\[\large -\ln|u|+C\]

BACK substituting into x gives us,\[\large -\ln|\csc x + \cot x|+C\]

Answer 6

Ok phew! We got through the first part ok.

Any confusion?

Answer 7

Looks good so far! haha never would have figured this out on my own.

Answer 8

Yah it's really really not obvious.
Requires a silly tricky at the start :\

Answer 9

So let's ummmm... ok ok let's try this..

Answer 10

In your second part, notice how the inside of that final log is csc x - cot x?
Remember how we ended up with ln|u| in the last one?

So my brain is thinking, if we multiplied by csc x + cot x and we ended up with ln|csc x+ cot x|, maybeeee if we multipy by csc x - cot x we'll end up with ln|csc x - cot x|!

Answer 11

\[\large \int\limits \csc x dx \left(\frac{\csc x - \cot x}{\csc x - \cot x}\right)\]

Answer 12

\[\large \int\limits \frac{\csc^2x-\csc x \cot x}{\csc x - \cot x}dx\]

Answer 13

\[\large \color{cornflowerblue}{u=\csc x - \cot x}\]Derivatives will give us,\[\large du=-\csc x \cot x +\csc^2x dx\]\[\large \color{cornflowerblue}{du=\csc^2x-\csc x \cot x dx}\]

Answer 14

Plugging in the blue pieces gives us,\[\large \int\limits \frac{du}{u}\]

Answer 15

And then similar steps to finish it off...
That worked out nicely huh? c:

Answer 16

Yeah, I will have to look over it a couple more times to fully understand it. But thank you you so much!

Answer 17

Yah also make sure you look up the derivatives for csc and cot :D
You could always convert to sines and cosines, but that's a big hassle.

Answer 18

OK, will do. :)