Answer:
We can use the identity ∫sec(x)dx = ln|sec(x) + tan(x)| along with integration by parts to evaluate the given integral.
Let u = sec(x) and dv = sec^2(x)dx. Then, du/dx = sec(x)tan(x) and v = tan(x).
By integration by parts, we have:
∫sec^3(x)dx = ∫sec(x)sec^2(x)dx
= sec(x)tan(x) - ∫tan^2(x)sec(x)dx
= sec(x)tan(x) - ∫sec(x)(sec^2(x) - 1)dx
= sec(x)tan(x) - ∫sec^3(x)dx + ∫sec(x)dx
Simplifying, we get:
2∫sec^3(x)dx = sec(x)tan(x) + ln|sec(x) + tan(x)| + C
Multiplying by 1/2, we get:
∫sec^3(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C
Now, let 8x = u, so that x = u/8 and dx = (1/8)du. Substituting, we get:
∫sec^3(8x)dx = (1/2)sec(8x)tan(8x) + (1/2)ln|sec(8x) + tan(8x)|/8 + C
Therefore, ∫sec^3(8x)dx = (1/2)sec(8x)tan(8x) + (1/16)ln|sec(8x) + tan(8x)| + C.
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