⌠ ⌡ cot x dx = | ⌠cos x ⌡sin x | dx |
set
u = sin x.
then we find
du = cos x dx
substitute du=cos x, u=sin x
| ⌠cos x ⌡sin x | dx = | ⌡u |
= ln |u| + C
substitute back u=sin x
= ln |sin x| + C⌠ ⌡ cot x dx = | ⌠cos x ⌡sin x | dx |
substitute du=cos x, u=sin x
| ⌠cos x ⌡sin x | dx = | ⌡u |
= ln |u| + C
substitute back u=sin x
= ln |sin x| + C