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July 13, 2014

July 13, 2014

Posted by **Brandon** on Sunday, June 10, 2012 at 12:05pm.

i know that cot^2x is csc^2(x)-1, but i just don't understand how to solve the cscx^(2/3), any help? i also know that its trig integrals/substitution...

- Calculus -
**Steve**, Sunday, June 10, 2012 at 5:24pmlet u = csc(x)

du = -csc(x) cot(x)

cot^2 = csc^2 - 1

and you have

csc^(2/3) (csc^2-1)cot(x)

= csc^(5/3)(csc*cot) - csc^(-1/3)(csc*cot)

= [-u^5/3 + u^(-1/3)] du

all downhill from there

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