Integral of csc(x), without that trick!

integral of csc(x) without that trick!,
integrals by student: Chase Snidfar,

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#blackpenredpen #math #calculus #apcalculus

[whiz]

Yeah yeah, that's all well and good, but can he hold 2 pens in one hand?

[avananana]

1:54 That face when you solve an integral after a few hours of hard work.

This is, in fact, a great video, never really thought you could break it down this "easily". I've only solved this integral with some weird four-way-inception substitution a long time ago. Might give it another shot in a while actually.

[francis]

Because you get 1/(u^2-1) after the substitution, isn't that also the same as -1/(1-u^2) meaning the integral of csc x can also be written as -tanh^-1(cos x) + c?

Just thought I'd point out an interesting relationship between the trig and the hyperbolic trig functions.

[MrDaveblev]

Use the power rule for logs on that last expression and you get ln ( sqrt( abs((cos x -1)/(cos x +1)))) which is the same as ln ( sqrt((1 - cos x)/(cos x +1))) which is = ln(tan(x/2)).

A little easier on the eyes.

[ostdog]

You could move the one half to the inside of the log and that makes it tan(x/2)

[matchamitminze]

Since this method involves memorization in some way (either by knowing that the antiderivative involves a log, so you’ll want to get a form of du/u, or just by rote trial and error), you can always rewrite csc(x) = 1/sin(x) = sin(x)/sin^2(x) = sin(x)/(1 - cos^2(x)) and setting u = cos(x), du = -sin(x)dx. From there, you can use partial fractions and simple algebraic manipulation with trig identities to arrive at the same answer. :)

[diego_fonseca]

No hablo Inglés pero he entendido a la perfección, muchas gracias!

[seegeeaye]

1/sinx = 1/(2sin(x/2) cos(x/2) = 1/2((sec(1/x))^2/(tan^x/2).
So we have the integral of d((tanx/2)/(tanx/2) = ln( (tan(x/2)) + C. We forget the "U business" and it makes simpler

[Drestanto]

2:00
Instead of pastial fraction decomposition, I like better to change 1/(u^2-1) into (1+u-u)/(1-u^2). It's also work, harder but trickier

[davidsojo]

Hi, thanks a lot for this video. I have one question though, when I solve the sum of the fractions, on the upper side I got -1

[phlaelogaming]

You look different in this video. You had a different hair cut here?

[cianmcgrath]

I used a different method. take u=cscx and use identities to rearrange equation in terms of u to -1/sqrt(u^2-1) and from there its easy to get -arcosh(cscx)

[kensonmalupande]

This is just a great explanation, keep it sir

[Miguel-xksl]

chase been smoking that stanky dank ganja green

[dildobaggins]

Dont know if its worth learning this way theres more to remember than the orginal method and its just a bit messy...

[craig]

Great video, been looking for something like this. I have a bigger expression sinxln(tanx))dx and I ended up with the integral of 1/sinx

[papajack]

Where's Steve? I believe his name was Steve.

[TheRedfire]

At 1:20 can you write the integral as (-1) (1/(1-u^2)) du put the constant (-1) outside the integral and substitute the integral for tanh^-1(u) ???

[TA]

Can't figure out how the presenter managed to do the partial fraction?

[bulldawg]

Good job, but now I have to look up the "cover up method" for partial fractions ... But dat OK ...