What is the integral of sec(1-z)tan(1-z) ?

The integral of sec(1-z)tan(1-z) is -sec(1-z)+C

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integral of sec(1-z)tan(1-z)

\int sec (1 - z) tan (1 - z) d z

- sec (1 - z) + C

Solution steps

\int sec (1 - z) tan (1 - z) d z

Apply u-substitution

: \int - sec (u) tan (u) d u

= \int - sec (u) tan (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= - \int 1 d v

Integral of a constant:

\int a d x = a x

= - 1 \cdot v

Substitute back

= - 1 \cdot sec (1 - z)

Simplify

= - sec (1 - z)

Add a constant to the solution

= - sec (1 - z) + C

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What is the integral of sec(1-z)tan(1-z) ?
The integral of sec(1-z)tan(1-z) is -sec(1-z)+C