What is the integral of sec^3(z) ?

The integral of sec^3(z) is 1/2 sec(z)tan(z)+1/2 ln|tan(z)+sec(z)|+C

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integral of sec^3(z)

\int sec^{3} (z) d z

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

Solution steps

\int sec^{3} (z) d z

Apply Integral Reduction

= \frac{sec ^{2} ( z ) sin ( z )}{2} + \frac{1}{2} \cdot \int sec (z) d z

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

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

Simplify

\frac{sec ^{2} ( z ) sin ( z )}{2} + \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

\frac{d}{d x} (sin (\frac{π x}{3}))

f (x) = ln (x^{2} + x + 1)

y^{^{''''}} - y = 0

\int x^{- 1} d x

\int_{0}^{2} e^{- t} d t

What is the integral of sec^3(z) ?
The integral of sec^3(z) is 1/2 sec(z)tan(z)+1/2 ln|tan(z)+sec(z)|+C