# Intergate 8 cot^3(x)?

asked Mar 19, 2013 in CALCULUS

integral 8 cot3(x)dx = 8  cot2(x)dx

integral 8 cot3(x) =dx 8  cot2(x) * cot(x) dx

Recall : Trigonometry formulas csc2(x) - cot2(x) = 1 Then csc2(x) - 1 = cot2(x)

Substitute cot2(x) = csc2(x) - 1   in the integral equation

integral 8 cot3(x)dx = 8  [(csc2(x) -1] * cotx dx

Recall : Distributive property (a - b) * c =(a*c) - (b*c)

integral 8 cot3(x) dx= 8 [(csc2(x) * cot(x) - 1 * cot(x)]dx

Recall : Commutative property of multiplication is a*b = b*a

integral 8 cot3(x) = 8 cot(x) * csc2(x) - cot(x)] dx

By the integral formula ∫[f(x) + g(x)] = ∫f(x) + ∫g(x)

integral 8 cot3(x) = 8 [[cot(x) * csc2(x) dx -  cot(x) dx]

put cot(x) = t Then dt = -csc2(x) dx There fore -dt = csc2(x) dx

Substitute cot(x) = t , csc2(x) dx = -dt in the [cot(x) * csc2(x) dx

integral 8 cot3(x) = 8 [ [t(-dt) -  cot(x) dx ]

integral 8 cot3(x) = 8 [- [tdt] -  cot(x) dx ]

integral 8 cot3(x) = 8 [-t2 / 2] -  cot(x) dx ]

Substitute t = cotx in the above equation

integral 8 cot3(x) = 8 [-cot2(x) / 2 -  cot(x) dx ]

Substitute int cot(x) dx = logsin(x) in the above equation

integral 8 cot3(x) = 8 [-cot2(x) / 2 - logsinx ]

Take out common term negitive one

integral 8 cot2(x) = -8 [(cot2(x)) / 2 + logsin(x)]