Question

use integration by parts to evaluate the integral: $int sin(ln(6x))dx$

Explanation:

Step1: Let \(u = \sin(\ln(6x))\) and \(dv=dx\)

First, find \(du\) and \(v\).
We know that if \(y = \ln(6x)\), then \(\frac{dy}{dx}=\frac{1}{x}\).
By the chain - rule, if \(u=\sin(\ln(6x))\), then \(du=\cos(\ln(6x))\cdot\frac{1}{x}dx\).
And if \(dv = dx\), then \(v=x\).

Step2: Apply integration - by - parts formula \(\int u\;dv=uv-\int v\;du\)

\(\int\sin(\ln(6x))dx=x\sin(\ln(6x))-\int x\cdot\frac{1}{x}\cos(\ln(6x))dx=x\sin(\ln(6x))-\int\cos(\ln(6x))dx\)

Step3: Apply integration by parts again for \(\int\cos(\ln(6x))dx\)

Let \(u = \cos(\ln(6x))\) and \(dv = dx\). Then \(du=-\sin(\ln(6x))\cdot\frac{1}{x}dx\) and \(v = x\).
\(\int\cos(\ln(6x))dx=x\cos(\ln(6x))+\int\sin(\ln(6x))dx\)

Step4: Substitute the result of \(\int\cos(\ln(6x))dx\) into the previous equation

\(\int\sin(\ln(6x))dx=x\sin(\ln(6x))-(x\cos(\ln(6x))+\int\sin(\ln(6x))dx)\)
\(\int\sin(\ln(6x))dx=x\sin(\ln(6x))-x\cos(\ln(6x))-\int\sin(\ln(6x))dx\)

Step5: Solve for \(\int\sin(\ln(6x))dx\)

Add \(\int\sin(\ln(6x))dx\) to both sides:
\(2\int\sin(\ln(6x))dx=x(\sin(\ln(6x))-\cos(\ln(6x)))\)
\(\int\sin(\ln(6x))dx=\frac{1}{2}x(\sin(\ln(6x))-\cos(\ln(6x)))+C\)

Answer:

\(\frac{1}{2}x(\sin(\ln(6x)) - \cos(\ln(6x)))+C\)