- #1

ago01

- 46

- 8

- Homework Statement
- ##\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx##

- Relevant Equations
- U-Sub

##\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx##

Let ##u = \csc{x}##

then

##-du = \csc{x}\cot{x}dx##

So,

##\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx##

##-\int \frac{1}{1+u^2}du = -\arctan{u} + C##

##-\arctan{\csc{x}} + C##

This answer was wrong. The actual answer involved fully simplifying the fraction and using

##u = \sin{x}##.

I understand how this approach is done and it makes complete sense.

What I'm confused about is where I went wrong in my derivation. I don't see my mistake, and the only difference is I ended up with ##\csc{x}## and a negative sign.

Could you help me spot my error? Usually bad u-subs makes things harder so I must've made a dumb subtle mistake I can't see.

Let ##u = \csc{x}##

then

##-du = \csc{x}\cot{x}dx##

So,

##\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx##

##-\int \frac{1}{1+u^2}du = -\arctan{u} + C##

##-\arctan{\csc{x}} + C##

This answer was wrong. The actual answer involved fully simplifying the fraction and using

##u = \sin{x}##.

I understand how this approach is done and it makes complete sense.

What I'm confused about is where I went wrong in my derivation. I don't see my mistake, and the only difference is I ended up with ##\csc{x}## and a negative sign.

Could you help me spot my error? Usually bad u-subs makes things harder so I must've made a dumb subtle mistake I can't see.