# Integration Question

climbhi
If you had say &int;cos4(x)dx according to my integration table in calc book this would be something nasty. Could you not say let u = sin5(x)/5 therefore du = cos4(x)dx and then &int;du = u = sin5(x)/5 + C. Is there something wrong with this. This technique would work on &int;x2 if you said let u = x3/3 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: &int;cosn(x)dx = [(cosn-1x)(sinx)]/n + [(n-1)/n]&int;cosn-2(x)dx