Trigonometric identities for antidifferentiation - products of sines and cosines

[Stick]

I'm finding this really challenging. I don't feel the textbook (Essentials Ex 7E) has explained the theory behind it all very well and neither has my teacher. I managed to wing my way through the exercise and only get stuck on one question (although the others did take a long time and many attempts to do) but I still don't feel very confident about it. Is there anything I can do? Any advice, tips, resources? :S

[Nagisa]

the book "stewarts calculus" covers this topic well in chapter 7, maybe you could give that a flick through to see if it clears things up. Diversity in your learning is always good.

[b^3]

It just takes a bit of getting used to and recognising what you have to work with. You can look for a few things, for example if you can reduce the case to a sine or a cosine square (or even powers generally), then you can use your double angle formulas for cosine to reduce it down to a trig of power 1. If you are working with tan's and something other than sine and cosines, work with your identities to see if you can convert them into sines and cosines, or look for something that will give you an equivalent expression that becomes easier to work with. Now when you say have, the sines or cosines raised to powers that are greater than one, if they are both even you can reduce them down using the cosine double angle formula (can get annoying with higher powers, end up with a lot of and and such, and may have to do it again when you get a square of say a ). If one is even and one is odd, try breaking the odd one up so say something like, where is odd will becomes , the second term there will be even. Now when that is multiplied by sines, you can use a substitution to get rid of that odd power there, and get it down to just even powers, and convert the rest of the expression into the same sine or cos using the pythagorean identity, giving you something easier to work with.

It's just a matter of getting used to it, look at what you can break it into, look for parts of the integral that when differentiated will get rid of another part (leaving to a sub), try your even power to cosine double angle rules ( and ), try your trig identities.

[Stick]

Thanks! I've devised a few routine approaches now, although there still is the odd one which manages to throw something completely different at you.

[lzxnl]

It just takes a bit of getting used to and recognising what you have to work with. You can look for a few things, for example if you can reduce the case to a sine or a cosine square (or even powers generally), then you can use your double angle formulas for cosine to reduce it down to a trig of power 1. If you are working with tan's and something other than sine and cosines, work with your identities to see if you can convert them into sines and cosines, or look for something that will give you an equivalent expression that becomes easier to work with. Now when you say have, the sines or cosines raised to powers that are greater than one, if they are both even you can reduce them down using the cosine double angle formula (can get annoying with higher powers, end up with a lot of and and such, and may have to do it again when you get a square of say a ). If one is even and one is odd, try breaking the odd one up so say something like, where is odd will becomes , the second term there will be even. Now when that is multiplied by sines, you can use a substitution to get rid of that odd power there, and get it down to just even powers, and convert the rest of the expression into the same sine or cos using the pythagorean identity, giving you something easier to work with.

It's just a matter of getting used to it, look at what you can break it into, look for parts of the integral that when differentiated will get rid of another part (leaving to a sub), try your even power to cosine double angle rules ( and ), try your trig identities.

I actually wouldn't recommend converting tangents to sines and cosines. You should commit these two formulas to memory:
d(sec x)/dx = sec x tan x
d(tan x)/dx = sec^2 x

With these, you can integrate any combination of sec x and tan x just as you would with sines and cosines.

Other than that, I agree with most of that. What you want is to either reduce everything down to sin or cos^2 kx, or you want to have sin x* (function of cosine) or cos x*(function of sin x) allowing for substitutions. This can be achieved if the degree of one of the sines or cosines is odd.
If they're both even, use the cosine double angle formula as mentioned.