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

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

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integral of (sec^3(x))

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

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

Solution steps

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

Apply Integral Reduction

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

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

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

Simplify

\frac{sec ^{2} ( x ) sin ( x )}{2} + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣ : \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

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

Add a constant to the solution

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

f (x) = x^{2} - 5 x + 5, (0, 5)

\frac{d}{d t} (e^{t} (cos (t)))

\frac{d}{d x} (x (ln (x) - 1))

\int_{0}^{\frac{3}{4}} \frac{1}{9 + 16 x ^{2}} d x

\frac{d}{d x} (\frac{1}{2 π} e^{- \frac{x ^{2}}{2}})

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