- #1

#### climbhi

^{4}(x)dx according to my integration table in calc book this would be something nasty. Could you not say let u = sin

^{5}(x)/5 therefore du = cos

^{4}(x)dx and then ∫du = u = sin

^{5}(x)/5 + C. Is there something wrong with this. This technique would work on ∫x

^{2}if you said let u = x

^{3}/3 and then did everything else the same except there its not quite so tricky. I guess what I'm asking is if you're good at designing a function that when differentiated would give the funtion in the integral can you use my method there instead of the tables which give this nasty formula: ∫cos

^{n}(x)dx = [(cos

^{n-1}x)(sinx)]/n + [(n-1)/n]∫cos

^{n-2}(x)dx