Question

the integral of csc^4 (x) cot^6 (x) dx

Answer 1

You know that the derivative of cotangent is negative cosecant squared, so I'd start off with breaking up the powers like so:
\[\csc^4x\cot^6x=\csc^2x\csc^2x\cot^6x\]
Now apply a trig identity to one of the csc terms:
\[\cdots=\csc^2x\left(1+\cot^2x\right)\cot^6x\]
Then distribute:
\[\cdots=\csc^2x\left(\cot^6x+\cot^8x\right)\]
So to integrate \(\displaystyle\int \csc^2x\left(\cot^6x+\cot^8x\right)~dx\), substitute \(u=\cot x\).

Answer 2

I may have gotten it
\[-\frac{ \cot ^{7} }{ 7 } - \frac{ \cot ^{9}}{ 9 }\] +constant

Answer 3

Yes, that's correct

Answer 4

Oh awesome! thank you for the help!

Answer 5

yw