use integration by parts to evaluate the integral: $int sin(ln(8x))dx$

Explanation:

Step1: Let $u = \sin(\ln(8x))$, $dv=dx$

$du=\frac{\cos(\ln(8x))}{x}dx$, $v = x$ By integration - by - parts formula $\int u;dv=uv-\int v;du$, we have $\int\sin(\ln(8x))dx=x\sin(\ln(8x))-\int x\cdot\frac{\cos(\ln(8x))}{x}dx=x\sin(\ln(8x))-\int\cos(\ln(8x))dx$.

Step2: For $\int\cos(\ln(8x))dx$

Let $u = \cos(\ln(8x))$, $dv = dx$. Then $du=-\frac{\sin(\ln(8x))}{x}dx$, $v = x$. By integration - by - parts formula, $\int\cos(\ln(8x))dx=x\cos(\ln(8x))+\int x\cdot\frac{\sin(\ln(8x))}{x}dx=x\cos(\ln(8x))+\int\sin(\ln(8x))dx$.

Step3: Substitute $\int\cos(\ln(8x))dx$ into the first result

$\int\sin(\ln(8x))dx=x\sin(\ln(8x))-(x\cos(\ln(8x))+\int\sin(\ln(8x))dx)$.

Step4: Solve for $\int\sin(\ln(8x))dx$

Let $I=\int\sin(\ln(8x))dx$. Then $I=x\sin(\ln(8x))-x\cos(\ln(8x)) - I$. $2I=x\sin(\ln(8x))-x\cos(\ln(8x))$. $I=\frac{x}{2}(\sin(\ln(8x))-\cos(\ln(8x)))+C$.

Answer:

$\frac{x}{2}(\sin(\ln(8x))-\cos(\ln(8x)))+C$