How do I integrate sec x * csc x?
I’m sure I should know how to do this. The answer seems to involve something like ln |csc 2x + cot 2x|. The answer’s not the important thing, but how to find the answer is.
Thanks.
How do I integrate sec x * csc x?
I’m sure I should know how to do this. The answer seems to involve something like ln |csc 2x + cot 2x|. The answer’s not the important thing, but how to find the answer is.
Thanks.
Oh, wait, never mind. I use the double angle formula to rewrite the denominator…
Does it help if you rewrite it as 1/(sin x * cos x) ?
I can’t explain the intuition, but looking at an old test of mine with this problem, I ended up doing
integral( 1/u du) where u = tan(x), du=(sec^2(x))
Then the integral of 1/u with respect to u is log(u), so you get log(tan(x)) (which you can reduce or change to your liking using properties of logs and trig identities).
Unfortunately, it’s been a few years since Integral Calc, so I honestly can’t remember WHY I did it this way, and to be embarrassingly honest, I forgot how to do the entire process of assigning du to a value to begin with.
ETA: I most likely used a table for some of the trig stuff, which is probably why I have trouble remembering.
In the end, every integration amounts to “Find something whose derivative (by straightforward, mechanical differentiation rules) is the integrand”. “Learning to integrate” is just learning a bunch of search heuristics on top of that for efficiency.
Still, in this case, let’s see how such a heuristic search might go:
First, rewrite the integrand as 2/sin(2x) dx by the double angle formula, as you’ve noted.
Then, do a “u substitution” (i.e., try supposing that there is a chain rule at the end of the differentiation we’re undoing): let u = 2x, so du = 2 dx. Our integrand is the integral of 1/sin(u) du = csc(u) du.
So if you know how to integrate 1/sin(u) du, you’re golden.
But maybe you don’t… so let’s search around some more.
Here’s the bit of cleverness: rewrite this as sin(u)/sin(u)^2 du = sin(u)/(1 - cos(u)^2) du. Do another chain rule inversion: let v = cos(u), so dv = -sin(u) du. Thus, we are integrating -1/(1 - v^2) dv.
Ok, so long as you know how to integrate -1/(1 - v^2) dv, you are golden
Don’t know how to integrate this? Alright, let’s keep going.
Rewrite this as -1/2 * [1/(1 + v) + 1/(1 - v)] dv.
Integrating this gets us -1/2 * [the integral of 1/(1 -v) dv + the integral of 1/(1 + v) dv]. So long as you know how to do the integral of 1/x dx, both of these integrals are easy.
And let’s cut this post short and assume you know that ln(x) is the integral of 1/x dx. So we end up with -1/2 * (ln(1 + v) - ln(1 - v)), where v = cos(u), where u = 2x. [All up to an additive constant, of course]
This isn’t the only nice form in which the answer could be expressed, mind you, but it is one perfectly reasonable one.
I would write it as d(sin x)/(sin x)(cos^2 x) = d(sin x)/(sin x)(1-sin^2x) = dt/t(1-t)(1+t) and write the denominator as a/t + b/(1-t) + c/(1+t) and the integral will come out as a(log t) + b(log(1-t)) + c(log(1+t)) + d, where, of course, t = sin x.