I am having difficulty finding integral((tanx)^2 / (secx)^2)dx.
I tried converting everything in terms of sinx and cosx but that doesn't seem to help.
Once you've simplified and gotten an expression in just sine, you can use the double-angle formula (in reverse) for the cosine:I am having difficulty finding integral((tanx)^2 / (secx)^2)dx.
I tried converting everything in terms of sinx and cosx but that doesn't seem to help.
I'm still not sure I understand. Once I convert this into terms of sinx and cosx I have integral((sinx)^2 / (cosx)^4)dx. After that I should use the double angle forumual or half angle? The one you showed me is the half angle formual.Once you've simplified and gotten an expression in just sine, you can use the double-angle formula (in reverse) for the cosine:I am having difficulty finding integral((tanx)^2 / (secx)^2)dx.
I tried converting everything in terms of sinx and cosx but that doesn't seem to help.
. . . . .
Then the integral is pretty straightforward.
Note: You will be using the double-angle formulas (in reverse) a lot in calculus, so make sure you're solid on them!