Question

Integral(csc^6(x)cot(x))dx

Answer 1

isnt the derivative of cosecant something like "cosecant cotangent"? or maybe its negative ?

Answer 2

yeah, its negative
so make \(u=\csc(x), du =-\csc(x)\cot(x)\) and get
\[-\int u^5du\]

Answer 3

I would start by simplifying this as much as possible. csc^6 is equal to 1/sin^6. Cotangent is equal to cos/sin.
\[\int\limits \frac{ 1 }{ \sin^6 x } \times \frac{ \cos x }{ \sin x } = \int\limits \frac{ \cos x }{ \sin^7 x }\] This may or may not help you, I'm just throwing this out there.

Answer 4

Satellite. This looks like where I was headed...

Answer 5

And look, it's a simple u sub. This is what I get for not memorizing those weird trig derivatives

Answer 6

same idea as @vinnv226

Answer 7

actually @vinnv226 there is not that much to memorize
if you know that the derivative of secant is secant tangent, then to take the derivative of the "co" function, replace all by "co" and stick a minus sign in front of it

Answer 8

for example
the derivative of tangent is secant squared, so the derivative of cotangent is minus cosecant squared

Answer 9

Actually you could probably do my way and make it work. What if we make this
\[\int\limits \cos x \sin^{-7} x\]
Let u= sin x
du=cos x dx
\[\int\limits u^{-7} du\]

Answer 10

yeah your way will work too for sure

Answer 11

So:
\[(-1/6)u^{-6} = -\sin^{-6} / 6\]

Answer 12

then when you get two different answers, you can figure out why they are actually the same

Answer 13

Trig integrals are always interesting

Answer 14

personally, i loath integration

Answer 15

And actually, thats exactly the same thing you get by doing @satellite73 s method above.

Answer 16

Thanks a bunch guys! Really helped