Question

1. $\int \sin 3x dx$, $u = 3x$

Explanation:

Step1: Differentiate \(u\)

If \(u = 3x\), then \(du=3dx\), and \(dx=\frac{1}{3}du\).

Step2: Substitute into the integral

\(\int\sin(3x)dx=\int\sin(u)\cdot\frac{1}{3}du\).

Step3: Integrate \(\sin(u)\)

We know that \(\int\sin(u)du=-\cos(u)+C\). So \(\frac{1}{3}\int\sin(u)du=\frac{1}{3}(-\cos(u)) + C\).

Step4: Substitute back \(u = 3x\)

\(\frac{1}{3}(-\cos(3x))+C=-\frac{1}{3}\cos(3x)+C\).

Answer:

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