Question

make the given substitution to evaluate the indefinite integral.

$$ int sin ( 8 x ) d x, quad u = 8 x $$

$$ int sin ( 8 x ) d x = - \frac { 1 } { 8 } cos ( 8 x ) + c $$

Explanation:

Step1: Differentiate \(u\)

Given \(u = 8x\), then \(du=8dx\), so \(dx=\frac{1}{8}du\).

Step2: Substitute into the integral

Substitute \(u = 8x\) and \(dx=\frac{1}{8}du\) into \(\int\sin(8x)dx\), we get \(\int\sin(u)\cdot\frac{1}{8}du\).

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

Since \(\int\sin(u)du=-\cos(u)+C\), then \(\frac{1}{8}\int\sin(u)du=\frac{1}{8}(-\cos(u)) + C\).

Step4: Substitute back \(u = 8x\)

Substitute \(u = 8x\) back, we have \(-\frac{1}{8}\cos(8x)+C\).

Answer:

\(-\frac{1}{8}\cos(8x)+C\)