Question

Calculus question. Integral and substitution @zepdrix

Answer 1

Oh this silly problem :)
lol ya i remember this

Answer 2

It's good that they gave you the multiplication step,
because that's not something you can easily come up with on your own.
It's a messy trick to making this one work.

Answer 3

\[\large\rm \int\limits \csc x\left(\frac{\csc x+\cot x}{\csc x+\cot x}\right)dx=\int\limits\frac{\csc x(\csc x+\cot x)}{\csc x+\cot x}dx\]You understand that only the numerator is getting this cscx, ya? :)

Answer 4

Is this correct? so far

Answer 5

Oooo no no no,
let's not do sines and cosines :OO

It might be right, but uhhh boy that's a lot of extra work

Answer 6

ok, lol. What's your method?

Answer 7

...

Answer 8

du = (-csc x cot x) + (sec x tan x), yes?

Answer 9

oh, then du = (-csc x cot x) + sec^2 x

Answer 10

@zepdrix is that ok?

Answer 11

heck i ruined
\[\csc x \times \frac{ \csc x - \cot x }{ \csc x - \cot x }\]

Answer 12

???

Answer 13

Well first multiply out your numerator so you have an idea of what's going on.\[\large\rm =\int\limits \frac{\csc x \cot x+\csc^2x}{\csc x+ \cot x}dx\]What you're doing is,
you're letting your denominator =u,
and something magical happens :)

Answer 14

Ah, I think I see where you're getting @zepdrix

Answer 15

@StudyGurl14 just an alternate method
so u will have
\[\int\limits\frac{ \csc x (\csc x - \cot x) }{ \csc x - \cot x } dx\]

Answer 16

du/dx = \(\Large -\csc x \cot x - (-\csc^2 x)=-\csc x \cot x + csc^2 x\)

Answer 17

Wait, hold on

Answer 18

Made a mistake there

Answer 19

Woops, we're using csc x + cot x for our u, right? :O
You silly billy

Answer 20

du/dx = \(\Large -\csc x \tan x + \csc^2 x\)

So dx = \(\Large\frac{1}{-\csc x \tan x + \csc^2 x}\)

yea?

Answer 21

tack on a du at the end

Answer 22

cot x --> -csc^2x, right?

Answer 23

\[\csc x - \cot x = u\]
so \[du = (-\csc x \cot x + \csc^2 x) dx\]

Answer 24

ugh, I keep doing that lol. Yeah

Answer 25

So that'd be negative, then you pull out the -1, and it changes to