$\int \sin (\ln (2 x)) d x$
Answer
Finally, we check our solution by differentiating it and seeing if we get back the original integrand. The derivative of \(-x \cos(\ln(2x)) + \sin(\ln(2x))\) is indeed \(\sin (\ln (2 x))\), so our solution is correct.
Step 1 ：First, we recognize that this is an integral problem involving a composite function. We can solve it using the method of substitution.
Step 2 ：Let \(u = \ln(2x)\). Then, \(du = \frac{2}{2x} dx = \frac{1}{x} dx\). So, \(x du = dx\).
Step 3 ：Substitute \(u\) and \(dx\) into the integral: \(\int \sin(u) \cdot x du\).
Step 4 ：Now, we can use integration by parts to solve this integral. The formula for integration by parts is \(\int u dv = uv - \int v du\). We let \(u = x\) and \(dv = \sin(u) du\). Then, \(du = dx\) and \(v = -\cos(u)\).
Step 5 ：Substitute \(u\), \(v\), \(du\), and \(dv\) into the formula: \(uv - \int v du = x(-\cos(u)) - \int -\cos(u) dx = -x \cos(u) + \int \cos(u) dx\).
Step 6 ：Solve the integral \(\int \cos(u) dx = \sin(u)\).
Step 7 ：Substitute back for \(u\): \(-x \cos(\ln(2x)) + \sin(\ln(2x)) + C\), where \(C\) is the constant of integration.
Step 8 ：So, the solution to the integral \(\int \sin (\ln (2 x)) d x\) is \(-x \cos(\ln(2x)) + \sin(\ln(2x)) + C\).
Step 9 ：Finally, we check our solution by differentiating it and seeing if we get back the original integrand. The derivative of \(-x \cos(\ln(2x)) + \sin(\ln(2x))\) is indeed \(\sin (\ln (2 x))\), so our solution is correct.
