INTEGRATION of sinx/root(36-cos^2x)?

Let u = (1/6)cosx
Then, du = -(1/6)sinx dx and cosx = 6u

∫ {sinx dx / √(36 - cos²x)}
=(-6) ∫ {(-1/6) sinx dx / √[36 - (6u)²]}
=(-6) ∫ {du / √[6² - (6u)²]}
=(-6) ∫ {du / 6√(1 - u)²]}
= -∫{du/ √(1 - u)²]}
= -arccosu + C
= -arccos(cosx/6) + C