[MUSIC].
Sometimes the best substitution to make isn't even visible until after we've
messed around with the integrand some how.
For example, what's the antiderivative of 1 over 1 plus cosine x?
Dx, what can you do? your first move might be to try the
substitution u equals cosine x, which case du is negative sine x dx, but
there's no visible sine x in the integrand.
Well, since we're talking about invisible substitutions, we should try to mess
around with the integrand. Instead of doing that, let's try a
different trick. Let's try to multiply 1 over 1 plus
cosine x by 1 minus cosine x divided by 1 minus cosine x.
This doesn't change the integrand at all, because this is just 1.
This trick makes a hitherto invisible substitution visible.
I'm getting ahead of myself a little bit. Let's first apply a trig identity.
Oh, this is 1 minus cosine x in the numerator divided by, that's 1 minus
cosine squared x, and then the trig identity is that 1 minus cosine squared
x, well that's sine squared x. So now I want to antidifferentiate 1
minus cosine x over sine squared x dx. Well even that isn't so great.
Let's split it up. Well then I get, that this is
antiderivative 1 over sine squared x dx minus the antiderivative of cosine x over
sine squared x dx. Now that first integral is one that I can
do. Rewrite it as the antiderivative of
cosecant squared x. You know, I just have to think, do I know
any function whose derivative is cosecant squared.
Yes, negative cotangent is an antiderivative of cosecant squared, of of
x. What about that other intergral?
Well I could read this as cotangent times cosecant and just recognize the
antiderivative that way. But to demonstrate the technique I can
also apply u substitution to that antidifferentiation problem so let's let
u equal sine x, and in that case, du is cosine x dx, which is great, because
that's the numerator there. So this this antidifferentiation problem
becomes what? This is the antiderivative of du over u
squared. And I just gotta think, how do I,
antidifferentiate u to the negative second power?
Well that, is by the power rule plus1 over u plus c.
Alright, If I differentiate one over u, that gives me negative 1 over u squared.
Okay, but I don't want my answer to be in terms of u, right, I want my answer to be
in terms of x. So this is negative cotangent x plus,
what's 1 over u, well that's 1 over sine x which, if I wanted to, I could write as
cosecant x plus C. Let's put it all together.
So what I'm claiming here is that the antiderivative of 1 over 1 plus cosine x
dx is negative cotangent plus cosecant, plus some constant.
If you have it already, I hope you're getting the idea that there's some really
clever things that you can try to make these substitutions work.