Integral of csc

[KindofSlow]

When I evaluated ∫ csc x dx, I used csc x – cot x for mysubstitution instead of csc x + cot x

So my final answer is ln│csc x – cot x │ + C
The correct answer is -ln│csc x + cot x │ + C

My answer seems to also be correct but I cannot prove it sonot sure what I’m missing.

Any insight greatly appreciated.

Thank you

 

[stapel]

When I evaluated ∫ csc x dx, I used csc x – cot x for mysubstitution instead of csc x + cot x

So my final answer is ln│csc x – cot x │ + C
The correct answer is -ln│csc x + cot x │ + C

My answer seems to also be correct but I cannot prove it...

Try working on showing the arguments are the same, after moving the -1 in front of the second log to the inside of that log as a power. A Pythagorean identity will be very helpful...

 

[KindofSlow]

Aha, I think this is what you meant for me to do.
ln│csc x – cot x │
= - ln│1 / (csc x – cot x) │
= - ln│(csc x + cot x) / ((csc x)^2 – (cot x)^2) │
= - ln│(csc x + cot x) / ((cot x)^2+1 – (cot x)^2) │
= - ln│(csc x + cot x) │

Thank you very much stapel, I would not have gotten thiswithout your hints.

 

[stapel]

Aha, I think this is what you meant for me to do.
ln│csc x – cot x │
= - ln│1 / (csc x – cot x) │
= - ln│(csc x + cot x) / ((csc x)^2 – (cot x)^2) │
= - ln│(csc x + cot x) / ((cot x)^2+1 – (cot x)^2) │
= - ln│(csc x + cot x) │

Thank you very much stapel, I would not have gotten thiswithout your hints.

You got it! Good work!

 

[HallsofIvy]

What I would do:

\(\displaystyle \int csc(x)dx= \int \frac{dx}{sin(x)}\)

Seeing an odd power of sine I would immediately think of a cosine substitution:
\(\displaystyle \int \frac{sin(x) dx}{sin^2(x)}= \frac{sin(x)dx}{1- cos^2(x)}\)

Let u= cos(x) so that du= -cos(x)dx. So the integral becomes \(\displaystyle \int \frac{-du}{1- u^2}= -\int \frac{du}{(1- u)(1+ u)}\) which is easy to integrate.