Could use some help with these:
1. The integral sin^6 x cos^3 x dx
2. The integral sin^2 x cos^4 x dx
Thanks!
Hello, coolio!
The first one is quite straight forward . . .
$\displaystyle 1)\;\;\int \sin^6\!x\cos^3\!x\, dx$
We have: .$\displaystyle \sin^6\!x\cos^2\!x(\cos x) \;=\;\sin^6\!x(1-\sin^2\!x)(\cos x) \;=\;(\sin^6\!x - \sin^8\!x)(\cos x)$
The integral becomes: .$\displaystyle \int(\sin^6\!x - \sin^8\!x)(\cos x\,dx)$
Let: $\displaystyle u \:=\: \sin x\quad\Rightarrow\quad du \:=\:\cos x\,dx$
Substitute: .$\displaystyle \int(u^6 - u^8)\,du$
. . . Got it?
Thanks Soroban, that makes a lot more sense. The case where sin and cos both have even powers seems to be giving me the most trouble still. I know it has something to do with the half angle formulas.