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## Homework Statement

Evaluate the indefinite integral.

∫sin(27t) (sec(cos 27t))^2 dt

## Homework Equations

## The Attempt at a Solution

u = cos27t

du = -27sin27t

dx = du / -27sin27t

∫sin(27t)*(sec(u))^2 * (du/-27sin27t)

(-1/27) ∫(sec(u))^2 du

(-1/27)*tan(u)

**= (-1/27)*tan(cos27t) + C**?

I am going on with this problem the wrong way?