Question

I'm flat out confused about how to know when to apply reduction formulas and how to integrate products of sine and cosine or tangent and secant at different powers. I know I need to use trig identities but I can't ever seem to get my head around what the goal is. I just see that they tell me if m or n is odd or real or both even nonnegative numbers for exponents but that doesn't really give me enough information. Can someone help?

Answer 1

\[\int\limits_{}^{}\sin^3(x)cos^{-1}(x)dx\] Here's an example of what I'm talking about.

Answer 2

for this example, using a substitution will work nicely

u = cos^2
du = -2sin*cos dx
\[\rightarrow -\frac{1}{2}\int\limits_{}^{}\frac{1-u}{u} du\]

Answer 3

I can help with any integration problems :D Just tag me or something.

Answer 4

Can someone prove the reduction formulas, assuming that n is a positive integer? For example, the sine reduction formula is this:
\[\int\limits_{}^{}\sin^nxdx=-\sin^{n-1}xcosx/n+(n-1)/n \int\limits_{}^{}\sin^{n-2}xdx\]

Answer 5

Let me write it down on paper to make sure I have it right in my head then i'll show you :D

Answer 6

Ready?

Answer 7

proof comes from using integration by parts
\[u = \sin^{n-1} x ................dv = \sin x\]
\[du = (n-1)\sin^{n-2} x \cos x ............v = -\cos x\]
\[\int\limits_{}^{}\sin^{n} x = -\sin^{n-1} x \cos x +(n-1)\int\limits_{}^{}\sin^{n-2} x \cos^{2} x\]
\[\int\limits_{}^{}\sin^{n} x = -\sin^{n-1} x \cos x +(n-1)\int\limits_{}^{}\sin^{n-2} x (1-\sin^{2} x)\]
\[\int\limits_{}^{}\sin^{n} x = -\sin^{n-1} x \cos x +(n-1)\int\limits_{}^{}\sin^{n-2} x -(n-1)\int\limits_{}^{} \sin^{n} x\]
\[n \int\limits_{}^{}\sin^{n} x = -\sin^{n-1} x \cos x +(n-1)\int\limits_{}^{}\sin^{n-2} x \]
\[ \int\limits_{}^{}\sin^{n} x =\frac{ -\sin^{n-1} x \cos x}{n} +\frac{(n-1)}{n}\int\limits_{}^{}\sin^{n-2} x \]