Integration of ∫sec(x)dx

There is no obvious way to integrate the secant function, sec(x). One way is to use the method of partial fractions. While this method takes a quite a few steps to solve the integral, it is robust. It does not rely heavily in the intuition of the student.

Another way involves some trickery where we try to form an integral of the form f'(x)/f(x), which gives the result ln|f(x)|. Although this method is efficient, it requires you to know part of the answer before you begin, which doesn't make much sense.

Using the method of partial fractions, the idea is to rewrite the integral of sec(x) as:

∫sec(x)dx = ∫cos(x)dx / (1 - sin^2(x))

Then by letting u = sin(x)...

∫sec(x)dx = ∫du / (1 - u^2) = ∫ du / (1+u)(1-u)

We can then convert the expression 1/(1+u)(1-u) into the sum of its partial fractions.

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[verride]

Finally a method that doesn't have me miraculously use sec x + tan x . Thank you!

[Paul-obhy]

at 3.40 instead of using partial fractions, that function is actually integratable

you get
tanh^-1(u)

[harshvardhan]

I am from India.your video is very useful for me.👍👍

[isaacnogueira]

Hello. I’m Brazilian and I don’t write ingles very well. But I want tell you that your video is very good, and it help me understand this integrate easily. Your explanation is simple and direct. Tank you. I put this link in my favorite and I signed up your channel. I’m sorry because I prefer write wrong instead use google translation

[ultraforte]

But we can also make a substitution like this: cos(x)=(1-t^2)/(1+t^2) where t=tan(x/2) and so on. These are the parametric equations of trigonometry.

[Silentstrike_]

Thank you for showing this! I've been stuck on the last steps of the first method for quite a while now - and all the other guides I've found use the (sec(x) + tan(x)) trick (the second method). The second method works great - but like you said, you'd need to know the answer to know how to do it, which defeats the point somewhat.

[lekesanusi]

The 2nd step is easier and less complicated. But we could as well get the value of dx. Which we can substitute. Then canceling will occur. There by replacing the bottom with U. And then we are left with integral 1/U which will = ln U + C. Finally replace U

[emmanuelalbazi]

Good job master.
I liked the first method.
I can do it with the integral of csc(x).

[thomasarch]

HERE IS AN EASIER WAY!! Multiply numerator and denominator by cos x and we have // (cos x)/(( cos x)^2) which is //(cos x) / (1 - (sin x)^2), Now let u = sin x and du = cos x dx
and we get // 1 / (1 - u^2) which is the standard integral for the inverse hyperbolic tangent function, so the answer is artanh( u) + or artanh (sin x) + c. Note I use a double slash “//“ to
denote the integral sign. This same approach can be used to solve the integral of the cosecant function which is minus artanh (cos x) + c.
You can also solve this using the Weirestrauss substitution method where we let tan (x/2) = z, sin x = 2z/(1+z^2). cos x = (1 - z^2)/ (1 + z^2), and dx = 2dz /(1 + z^2) so in this case we have
// 2 / ( 1 - z^2) dz = 2 artanh (tan x/2) + c. The way shown here in this video is usually the first one taught in a Calculus I course but requires you to “know” part of the answer before you start, kind of like guessing?

[wiltherdelacuesta]

NIce excercise by solving it with partial fraccions, Thay way you keep working on your algebra knowlege which is funadamental to solve integration problems...the substitusion way is more a solution for dummies;))

[robertalkemade]

well done
partial fractions in precise and simple steps
thanks again

[MisterTutor]

I going hazard a guess that the integral of csc x is ln |csc x + cot x| + C. The calculus formulas do have a certain symmetry for lack of a better term.

[_muz_]

Simply multiply and divede by secx+tanx and let secx+tanx = t u will get answer in 5 lines

[princessokelu]

at 4:17 the numerators aren't both 1/2 one is suppose to -1/2 and the other is suppose to be 1/2

[moaadmaaroufii]

1/cosx = cosx/2(1-sinx) +cosx/2(1+sinx)

[lecinquiemeroimage]

And what about this one : ∫ ln (cosx) dx ? (not so easy ...)
You can also try to calculate ∫₀ₐ ln (cos x) dx (so x between 0 and a, with a = π/4).

[tonymunene]

Great explanation but at 5.31..why does the sign change to minus

[thatoneophiucus]

How do you know when/whether to multiply by the numerator or the denominator?

[sumantchopde]

How come there's a minus at 5:22 instead of + between those two terms?

[austinnewton]

I don't understand how you get 1/2 for the numerators of the partial fractions. what do you mean "you happen to know?"