Question

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

Explanation:

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

$du=\frac{\cos(\ln(8x))}{x}dx$, $v = x$

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

$\int\sin(\ln(8x))dx=x\sin(\ln(8x))-\int x\cdot\frac{\cos(\ln(8x))}{x}dx=x\sin(\ln(8x))-\int\cos(\ln(8x))dx$

Step3: For $\int\cos(\ln(8x))dx$, use integration by parts again. Let $u = \cos(\ln(8x))$ and $dv = dx$

$du=-\frac{\sin(\ln(8x))}{x}dx$, $v = x$

Step4: Apply integration - by - parts formula for $\int\cos(\ln(8x))dx$

$\int\cos(\ln(8x))dx=x\cos(\ln(8x))+\int x\cdot\frac{\sin(\ln(8x))}{x}dx=x\cos(\ln(8x))+\int\sin(\ln(8x))dx$

Step5: Substitute $\int\cos(\ln(8x))dx$ into the result of Step 2

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

Step6: Solve for $\int\sin(\ln(8x))dx$

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

Answer:

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