What is the integral of tan^2(z)sec(z) ?

The integral of tan^2(z)sec(z) is 1/2 sec(z)tan(z)-1/2 ln|tan(z)+sec(z)|+C

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integral of tan^2(z)sec(z)

\int tan^{2} (z) sec (z) d z

\frac{1}{2} sec (z) tan (z) - \frac{1}{2} ln ∣ tan (z) + sec (z) ∣ + C

Solution steps

\int tan^{2} (z) sec (z) d z

Rewrite using trig identities

= \int (- 1 + sec^{2} (z)) sec (z) d z

Expand

(- 1 + sec^{2} (z)) sec (z) : - sec (z) + sec^{3} (z)

= \int - sec (z) + sec^{3} (z) d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int sec (z) d z + \int sec^{3} (z) d z

\int sec (z) d z = ln ∣ tan (z) + sec (z) ∣

\int sec^{3} (z) d z = \frac{1}{2} sec (z) tan (z) + \frac{1}{2} ln ∣ tan (z) + sec (z) ∣

= - ln ∣ tan (z) + sec (z) ∣ + \frac{1}{2} sec (z) tan (z) + \frac{1}{2} ln ∣ tan (z) + sec (z) ∣

Simplify

- ln ∣ tan (z) + sec (z) ∣ + \frac{1}{2} sec (z) tan (z) + \frac{1}{2} ln ∣ tan (z) + sec (z) ∣ : \frac{1}{2} sec (z) tan (z) - \frac{1}{2} ln ∣ tan (z) + sec (z) ∣

= \frac{1}{2} sec (z) tan (z) - \frac{1}{2} ln ∣ tan (z) + sec (z) ∣

Add a constant to the solution

= \frac{1}{2} sec (z) tan (z) - \frac{1}{2} ln ∣ tan (z) + sec (z) ∣ + C

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What is the integral of tan^2(z)sec(z) ?
The integral of tan^2(z)sec(z) is 1/2 sec(z)tan(z)-1/2 ln|tan(z)+sec(z)|+C