Question

what is the integral of csc^8 6x dx?

Answer 1

You can rewrite the integrand using the Pythagorean identity \(\bigg(\csc^2x=1+\cot^2x\bigg)\):
\[\begin{align*}\large\int\csc^8(6x)~dx&=\large\int\csc^2(6x)\left(\csc^2(6x)\right)^3~dx\\
&=\large\int\csc^2(6x)\left(\bigg(1+\cot^2(6x)\bigg)\right)^3~dx\end{align*}\]
First, let's get rid of that \(6x\). Let \(u=6x\), so you have \(du=6~dx,\) or \(\dfrac{1}{6}du=dx\).
\[\large\frac{1}{6}\int\csc^2u\left(\bigg(1+\cot^2u\bigg)\right)^3~du\]
Let \(t=\cot u\), so that \(dt=-\csc^2 u~du\), or \(-dt=\csc^2 u~du\).
\[\large-\frac{1}{6}\int\left(1+t^2\right)^3~dt\]
Expand the denomintor:
\[(1+t^2)^3=1+3t^3+3t^4+t^6\]
Integrate term by term:
\[\large\int\left(1+3t^3+3t^4+t^6\right)~dt\]
Should be a piece of cake from here.