integragtionxcosx dx/(sin^2 X)?

Question

Integration of xcosx /(sin^2 X)

Answer 1

write the integral as
x cot(x)csc(x) dx

use integrationby parts
let u = x, so du = dx
let dv = cot(x)csc(x), so v = -csc(x)

so the integral is
uv - int(vdu)
-xcsc(x) + int(csc(x)dx)

to find int(csc(x)dx) multiply csc(x) by csc(x) - cot(x) / [(csc(x) -cot(x)]

so you have csc²(x) - cot(x)csc(x) / [csc(x) - cot(x)]
let w = csc(x) - cot(x)
dw = -csc(x)cot(x) + csc²(x) dx
so the integral is
int(dw/w) = ln|w| = ln|csc(x) - cot(x)|

answer is
-xcsc(x) + ln(csc(x) - cot(x)| + C

Answer 2

Call your integral I.
Integrate by parts:
Let
U = x dV = cos x dx /sin² x.
Then
dU = dx V = -1/sin x = -csc x.
So I = -x csc x + ln| csc x -cot x| + C.