[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.