Calculus help... Integrating with ln, trig, and subsitution?

Question

((1-cosx)/sinx)dx... I get that u is sinx and -du is cosx... but that leaves (1-du)/u... or maybe I am doing it wrong.

Answer 1

Because there is only one denominator in this problem, you might consider splitting the problem into two separate integrals. That is, the first integral containing (1/sinx) minus the second integral of (cosx/sinx). The first integral can be changed to cscx because (1/sinx) = cscx, so you're just finding the integral of cscx. The second integral is a basic u-sub where u=sinx, du=cosx, so you would get the integral of (du/u) which is lnu and replacing the the u with sinx, you obtain ln(sinx). You could also bypass the u-sub by changing (cosx/sinx) to cotx and finding the integral of that.

Hope this helps.

BTW: If the poster below is saying no to me, then she is saying no to the process she has taken as well. ROFL if that what she's saying no to.

Answer 2

No...just split the fraction up.
int: (1 - cos x) / sin x dx

Split out the fraction:
int: (1 / sin x) - (cos x / sin x)
int: csc x - cot x dx

Since you're subtraction, you can do the integral of csc x alone, then the integral of cot x alone.
int: csc x = ln |cosec x + cotan x| + C
int: cot x = ln |sin x| + C

Total: ln |cosec x + cotan x| - ln |sin x| + C