91Ó°ÊÓ

Since we know that \(\int \csc^2 x dx = - \cot x + C\), where C is an integration constant, we will use this rule to integrate \(\csc^2 6x\) as follows: $$\int \csc^2 6x dx = -\frac{1}{6} \cot 6x + C$$ Here, we factored out the constant '6' to follow along with the chain rule.

To check our result, we will differentiate the antiderivative with respect to \(x\), and see if we obtain the original function, \(\csc^2 6x\). Using the chain rule for differentiation, we have: $$\frac{d}{dx} \left[-\frac{1}{6} \cot 6x + C\right] = -\frac{1}{6} \frac{d}{dx} \left[\cot 6x\right]$$ Now, we know that the derivative of \(\cot x\) is \(-\csc^2 x\), so the derivative of \(\cot 6x\) will be \(-\csc^2 6x \cdot 6\) (applying the chain rule). Thus, we can write: $$-\frac{1}{6} \cdot (-\csc^2 6x \cdot 6) =\csc^2 6x$$ This shows that the differentiation of our antiderivative does give us the original function, \(\csc^2 6x\). Therefore, our answer is correct: $$\int \csc ^{2} 6x dx = -\frac{1}{6}\cot 6x + C$$