integral of sec(x), 4 results!

integral of sec(x), 4 results,

4 ways to integrate sec(x),
integral of 1/cos(x),
integral of sec(x) without that trick,

blackpenredpen,
math for fun,

[chrysophylaxs]

You could do
Int(secx) dx
Let u = secx
du = secx tanx dx
If you now form a right angled triangle with angle x, the hypotenuse will be u and the adjacent side will equal 1.
By the pythagorean theorem, the opposite side will be sqrt(u^2-1).
That means:
du = u sqrt(u^2-1) dx
For the cancellation:
dx = du/(u sqrt(u^2-1))
Now substitute
= Int(u · 1/(u sqrt(u^2-1))) du
The u and 1/u cancel to get:
= Int(1/sqrt(u^2-1)) du
Which is the inverse hyperbolic cosine
= cosh^-1(u)
Substitute back for u
= cosh^-1(secx)

[domthehypocrite]

If it were on a test I would go with #2 but if I knew I did really well on the test than I would go with #4 just to give my teacher a challenge in the midst of grading papers.

[iaagoarielschwoelklobo]

Jumpscare list:

3:25 *monster appears
19:19 *monster appears





btw nice vid

[WildSeven]

I liked the partial fraction solution most. Felt more intuitive and simple. cos^2x+sin^2x=1 is an old friend by now!

[ktuluflux]

You can also use the tangent half angle substitution. Also (erroneously) called the Weierstrass substitution, you can write any trig function in terms of tan(x/2)

There is a general case for integral (1/(a+bcosx)). Just put b = 1 and a= 0 in the final answer there.

[chrysophylaxs]

Now you could also write secx as cscx times tanx
That will get you
Int(cscx tanx) dx
Now let
u = tanx
du = sec^2(x) dx
Do the same trick again,
create a right-angled triangle with acutw angle x. The opposite side will be equal to u, and the adjacent side will be equal to 1. By Pythagoras, the hypotenuse is equal to sqrt(u^2+1).
Now, sec^2(x) = u^2+1
cscx = sqrt(u^2+1)/u
dx = du/(u^2+1)
Now substitute:
= Int( sqrt(u^2+1)/u · u · du/(u^2+1))
1/u and u will cancel:
= Int( sqrt(u^2+1)/(u^2+1) ) du
= Int( 1/sqrt(u^2+1) ) du
Which is the inverse hyperbolic sine!
= sinh^-1(u)
Substitute back for u
= sinh^-1(tanx) + C

[jimstiles]

I wish YouTube was around in 1987 when I was taking inviscid fluid flow. These videos are an excellent refresher in complex numbers.

[RyanLucroy]

If you love doing integrals the hard way, you can do the following:

Take the second method and replace the 1 in the nominator with sin^2(x)+cos^2(x).

After a lot of work you'll end up with
ln(sin(x/2)+cos(x/2)) - ln(cos(x/2)-sin(x/2))

Don't ask me how to get there, a friend of mine told me that :)

[victorlacerda]

The complex result is by far the best, thanks for all.

[Kuratius]

Another cool thing: the same approach as in #4 also works for the integral of 1/sin x :
you get -2*artanh (e^ix) as the antiderivative. I hadn't thought about that, thank you!

[mythicmansam]

Hey blackpenredpen I love your videos, I check everyday to see if you posted new content. The types of math problems you walk us through tickles my brain such an awesome way! Btw method number 4 was by far my favorite. It makes me wonder what other types of intergrals can have complex results :)

[utuberaj]

Very nice derivation, especially the last one using i and the one using stadard hyperbolic tanx.
But there is one more method in text books using the substitution u=tan(x/2), which reduces the trig function into rational function and is integrated in the stadard way we use for this class of functions. The result is I think = ln | tan(x/2+π/4) |+C.
This method is I learnt recently known as WEIRSTRASS substitution.

[OCinTexile]

I was a math geek in high school and college, but in my work I ended up never needing anything much beyond algebra and probability, never the calculus that I spent years learning, except to tutor a friend for a year at work who was having trouble with basic calc derivatives, and basically forgot all about it (that was 30+ years ago, even the tutoring was 20 years ago). I never suspected how much I still love this stuff until I started watching your videos (and Mathologers too) a couple months ago. I always try to solve your integrals, but am almost never able to - I don't even remember if we ever did integrals like the ones you show (and I went to a hard core math/science/engineering institute in California, studying calculus and advanced math for four years, plus a couple years in high school), and I am always amazed and binge watch your videos for hours on end. I'll never use it, but the pure beauty of it outweighs any practical use it might have or not. A BIG THANKS!

[serouj]

While you're at 6:38, you could have noticed it is in the form 1/1-k with k=u^2, you can turn that to an infinite series
in this case, sum (u^2n) {n, 0, infinity}
After integrating, it would be sum ((1/(2n+1))u^(2n+1)){n, 0, infinity}, then replace u by sin(x) which gives sum ((1/(2n+1))sin^(2n+1)x)){n, 0, infinity}+c

[PackSciences]

I prefer number 2 but number 4 is interesting to know.

[thephysicistcuber]

Use parametric formulas to express sec(x) in terms of tan(x/2), tan(x/2)=u, integrate

[Green_Eclipse]

If you stop method 4 at 2/i * integral of 1/(u^2+1)du then that equals 2/i * integral of 1/(u^2-i^2)du. Factor denomintor with difference of squares and use partial fraction. You end up with ln((e^(i*x) + i)/(e^(i*x) - i)) + C

[farukhsandhu]

Ur simile after solving a problem is great source of satisfaction😍

[imme]

We can write 1/cos(x) as tan(x)/sin(x), by putting u = sin(x) we get sqrt(1-u^2) = cos(x) and therefore
tan(x) = u/sqrt(1-u^2), and dx = du/sqrt(1-u^2). By simplyfing with "u", we get the integral of 1/[sqrt(1-u^2)*sqrt(1-u^2)], which is exactly what was found with the partial fraction or the inverse hyperbolic tangent

[Thalesfreitas]

Got a little heart attack from the mask...