Math Snap

STEP 1

1. The integral involves the functions $\sec(1-z)$ and $\tan(1-z)$, which are trigonometric functions.
2. We need to recognize that the integral of $\sec(u) \tan(u)$ with respect to $u$ is a standard integral.
3. We'll use a substitution method to simplify the integral.

STEP 2

1. Perform a substitution to simplify the integral.
2. Integrate with respect to the new variable.
3. Substitute back to the original variable.

STEP 3

Let $u = 1 - z$. Then, $du = -dz$.
$$dz = -du$$

STEP 4

Substitute $u = 1 - z$ and $dz = -du$ into the integral.
$$\int \sec(1-z) \tan(1-z) \, dz = \int \sec(u) \tan(u) \, (-du)$$

STEP 5

Rewrite the integral with the negative sign factored out.
$$\int \sec(u) \tan(u) \, (-du) = -\int \sec(u) \tan(u) \, du$$

STEP 6

Recognize that the integral of $\sec(u) \tan(u)$ with respect to $u$ is a standard integral.
$$\int \sec(u) \tan(u) \, du = \sec(u) + C$$

STEP 7

Apply the result to our integral.
$$-\int \sec(u) \tan(u) \, du = -\sec(u) + C$$

Substitute back $u = 1 - z$.
$$-\sec(u) + C = -\sec(1 - z) + C$$ Therefore, the solution to the integral is:
$$\int \sec(1-z) \tan(1-z) \, dz = -\sec(1 - z) + C$$