To solve the integral of \( t \cdot \sec^2(8t) \), we use integration by parts.

We let:
- \( u = t \), which gives \( du = dt \),
- \( dv = \sec^2(8t) \, dt \), which means \( v = \frac{1}{8} \tan(8t) \) (since the integral of \( \sec^2(8t) \) is \( \frac{1}{8} \tan(8t) \)).

Now apply the integration by parts formula:

\[
\int u \, dv = u \cdot v - \int v \, du
\]

Substituting in the values for \( u \) and \( v \):

\[
\int t \cdot \sec^2(8t) \, dt = t \cdot \frac{1}{8} \tan(8t) - \int \frac{1}{

Learn how to compute the integral of t(sec^2(8t)) using the method of integration by parts. This step-by-step solution covers key calculus concepts and involves trigonometric integration, suitable for undergraduate-level students.

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