∫[1/6,1/2] cscπt cotπt dt

Let u = πt

du = π dt

dt = (1/π) du

∫ cscπt cotπt dt = (1/π) ∫ csc u cot u du

= -(1/π) cot u

= -(1/π) csc πt

Let F(t) = -(1/π) csc πt

substitute the upper limit

F(1/2) = -(1/π) csc (π/2) = -(1/π)

substitute the lower limit

F(1/6) = -(1/π) csc (π/6) = -(2/π)

F(1/2)-F(1/6) = -(1/π) - ( -(2/π)) = 1/π

u=pit

du=pidt

I/pidu=dt

1/piintegral[cscucotudu]=

-1/pi(cscpi/2-cscpi/6)=-1/pi[1-2]=1/pi

cscpit = 1/(sinpit)

cotpit = cospit/sinpit

INT[(cospit/(sin^2pit) dv]

u substituion

u=sinpit

du= picospitdt

du/pi=cospitdt

substitute.

Answer: -1/(sinpit) + C

or

-cscpit + C.