Question

@satellite73 @solomonzelman

Answer 1

@SolomonZelman @jim_thompson5910

Answer 2

\(\color{#000000}{ \displaystyle \int \frac{1+\cos x}{\sin x} dx }\)

\(\color{#000000}{ \displaystyle \int \frac{1}{\sin x} dx +\int \frac{\cos x}{\sin x} dx }\)

or do you want an easier approach?

Answer 3

you don't get a notification unless you post...which you just did lol @chosenmatt

Answer 4

Whatever works is fine.

Answer 5

Yeah, the integral for csc(x), and for tan(x)....

Answer 6

oh, duh

Answer 7

cos/sin = cot, not tan

Answer 8

yes, thanks jim.
What the hack am I saying? :)

Answer 9

wait, I don't know the integral for csc and cot...?

Answer 10

My book only has for csc^2

Answer 11

We can proof the formulas for them, or you can just use the formula for them/

Answer 12

it is just the matter of willingness.

Answer 13

since I have to get my homework done soon, I'm gonna go with formula. Then, after, you can show proof

Answer 14

if it was me, i would look in the back of the book
almost all of calc 2 is right there on two pages

Answer 15

If it was me, I would just use mathematica))

Answer 16

lol or wolframalpha (in the browser already)

Answer 17

the book says the integral of cot is ln|sin x| + C

Answer 18

I checked the back

Answer 19

that one is a simple u - sub

Answer 20

oh, right

Answer 21

it is \[\int \csc(x)dx\] you need to look up

Answer 22

ok

Answer 23

-ln|csc x - cot x| + C

Answer 24

Is that right?

Answer 25

\(\color{#000000}{ \displaystyle \int \frac{1+\cos x}{\sin x} dx }\)

\(\color{#000000}{ \displaystyle \int \frac{2+e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}} dx }\)

yeah, either way you end up with the same thing, although in a bit different form:)

Answer 26

You are right about the ln

Answer 27

got my book right here, let me check

Answer 28

You can get that by multiplying the numberator and denominator by csc+cotx and negating the top to get integral of -1/u form

Answer 29

Integral of cot(x)dx is simplier.

Answer 30

my book does not have the minus sign

Answer 31

weird? mine does...?

Answer 32

\[\ln(\csc(x)-\cot(x))\]

Answer 33

maybe it doesn't matter...?

Answer 34

possible
we can check by differentiation

Answer 35

Idk why I even have to use that formula when I haven't even learned it yet

Answer 36

It does and there should be a minus

Answer 37

that is the point of a formula
you don't have to learn it, just look it up

Answer 38

-ln abs(cotx+cscx)

Answer 39

maybe my Stewart has a typo
i think he just passed away

Answer 40

Why it's important to recognize the steps needed for that result

Answer 41

\[\log(\csc(x)-\cot(x))\] is definitely right
just check it

Answer 42

??

Answer 43

multiply cscxdx by cotxcscx / cotxcscx you get csc^2+cotxcscx / cotx+cscx yeah?

Answer 44

right?

Answer 45

The first time I integrated \(\csc x\) in my life, I remember, I ended up with.

\(\color{#000000}{ \displaystyle \int\frac{1}{\sin x}dx =\int\frac{\sin x}{1-\cos^2x}dx }\)

Sub,
\(u=\cos x\)
\(du=-\sin x dx\)

\(\color{#000000}{ \displaystyle -\int\frac{1}{1-u^2}du=\int\frac{1}{u^2-1}du }\)

Partial Fractions,
\(\displaystyle \frac{1}{u^2-1}=\frac{A}{u+1}-\frac{B}{u-1}\)

\(\displaystyle 1=A(u-1)-B(u+1)\)

\(\displaystyle B=-1/2\)
\(\displaystyle A=1/2\)

\(\color{#000000}{ \displaystyle \int\frac{1}{u^2-1}du=\int\frac{1/2}{u+1}du-\int\frac{1/2}{u-1}du = }\)

\(\color{#000000}{ \displaystyle \frac{1}{2}\ln(\sin x+1)-\frac{1}{2}\ln(\sin x-1) = }\)

\(\color{#000000}{ \displaystyle \frac{1}{2}\ln\left(\frac{\sin x+1}{\sin x-1}\right)+C }\)

Answer 46

it is not very hard to proof these, just using u-substitution and an algebraic technique of partial fractions.

Answer 47

Oh, I left of absolute value. Sorry.

Answer 48

Good luck with integration:)

Answer 49

And it u=cosx so your answer should involve cosine

Answer 50

Is the answer ln|1-cos x| + c?

Answer 51

Yes, cosine instead of sine....

Answer 52

I overworked myself. gtg

Answer 53

lol, thanks

Answer 54

\[\int\limits cscx dx = \int\limits - (-\csc^2(x)-\cot(x)\csc(x))/\cot(x)+\csc(x) = \int\limits \frac{- 1 }{ u}du= -\ln \left| \cot(x) +\csc(x) \right| +C\]

Answer 55

u= cot(x) + csc(x)
du= -csc^2(x) - csc(x)cot(x)

Answer 56

So hopefully this convinces you that -ln... is the solution