Question

evaluate the indefinite integral. \\(\int \frac{\sin\left(\frac{8}{x}\
ight)}{x^2} dx = \square + c\\).

Explanation:

Step1: Substitution

Let \(u = \frac{8}{x}\), then \(du=-\frac{8}{x^{2}}dx\), and \(\frac{1}{x^{2}}dx =-\frac{1}{8}du\).

Step2: Integrate

The integral \(\int\frac{\sin(\frac{8}{x})}{x^{2}}dx\) becomes \(-\frac{1}{8}\int\sin(u)du\).
Since \(\int\sin(u)du=-\cos(u)+K\) (where \(K\) is a constant), then \(-\frac{1}{8}\int\sin(u)du=\frac{1}{8}\cos(u)+C\).

Step3: Back - substitution

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

Answer:

\(\frac{1}{8}\cos(\frac{8}{x})\)