What is the anti derivative of cscxcotx?

Answer 1

∫cscxcotx*dx

∫csc(u)cot(u)*du= -csc(u)+C, where C is the constant of integration

because d/dx(csc(u))=-[csc(u)cot(u)],

so d/dx(-csc(u))=csc(u)cot(u).

∫cscxcotx*dx

Let:

u=x

du/dx=1

du=dx

∫cscucotu*du= -csc(u)+C

Plug in x for u.

∫cscxcotx*dx= -csc(x)+C