Evaluate this question integral of {cotx/log(sinx)dx}?

 I = ⌡ {cotx / log(sinx) } dx

Let u=log(sinx)

du/dx = (1/sinx) * d(sinx)/dx

du/dx = cosx/sinx

du = cotx dx

Therefore : I = ⌡ {1 / u } du

I = log(u) + C

I = log(logIsinxI)+C

Regards

Arun (askIITians forum expert)