7.2 - Trigonometric Integration

Last time (on Dragonball Z...) we learned a neat little formula that we could substitute parts in to help evaluate more complicated integrals, and God bless whoever figured that out. There are still some integrals, mainly with more complex trig functions, that don't quite work. For example, if we have
⌠cos³x dx, then the only real substitution we could make would be to set u to cos x. If we did that, our du would be -sinx, which doesn't help us because we don't have one in there to get a du out of. If we want to integrate with cosine, then we have to have a sine function in there to make our du work, and vice versa with integrating a sine function.
But, being the clever mathematicians we are, we can take out of the cosine functions and call it cos²x(cos x). We know that cos²x = 1-sin²x, so this becomes (1-Sin²x)(cos x)dx. Now that we have sines and cosine, we can set u=sin x and get (1-u²)cos x dx. Now that our du is cos x, we get ⌠(1-u²)du, which is perfectly doable.

Since, as I said in the last entry, I'm pressed for time and need to get on with it, I'm going to jump straight to the strategy summaries.

For sine and cosine functions:
1. If the power of cosine is odd, save one of them out and change the remaining even powers of cosine to powers of (1-sin²x), set your u to sin x and du to cos x, and go from there.
2. If the power of sine is odd, save one of them and change the remaining to (1-cos²x) and do the same as above.
3. If both of them are even, use the half-angle identities instead in whatever path you feel works best.

For tangent and secant functions:
1. If the power of secant is even, save a factor of sec²x (make sure you're saving a squared factor, not just sec x, which isn't terribly useful) change the rest to a power of (1+tan²x), with u as tan²x and du as sec²x.
2. If the power of tangent is odd, pull out a (sec x)(tan x) then use (sec²x-1) for the rest of them, then set your u to sec x and tan x as your du.

For products of sine and cosine functions with constants A and B:
1. sinAcosB = .5[Sin(A-B) + sin(A+B)
2. sinAsinB = .5[cos(A-B) - cos(A+B)]
3. cosAcosB = .5[cos(A-B) + cos(A+B)]

The last three weren't proven in the textbook, but he promises that you get that if you solve with integration by parts, so I'll just take his word for it.

That's all for now. I might go back into more depth later, but I have a crapton to do.