Question

1. $\int\frac{dx}{\sin^{2}3x}$

Explanation:

Step1: Use trigonometric identity

Recall that \(\frac{1}{\sin^{2}u}=\csc^{2}u\), so \(\int\frac{dx}{\sin^{2}3x}=\int\csc^{2}(3x)dx\).

Step2: Use substitution

Let \(u = 3x\), then \(du=3dx\), and \(dx=\frac{1}{3}du\). The integral becomes \(\frac{1}{3}\int\csc^{2}u du\).

Step3: Integrate \(\csc^{2}u\)

We know that \(\int\csc^{2}u du=-\cot u + C\). So \(\frac{1}{3}\int\csc^{2}u du=-\frac{1}{3}\cot u + C\).

Step4: Substitute back

Substitute \(u = 3x\) back, we get \(-\frac{1}{3}\cot(3x)+C\).

Answer:

\(-\frac{1}{3}\cot(3x)+C\)