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

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

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integral of sec^3(u)

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

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

Solution steps

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( 16 + x ^{2} ) ^{3}} d x

- 6 x = 10 - y - 36 x = 48 - 6 y

\int 4 cos^{2} (x) tan^{3} (x) d x

\frac{\partial}{\partial x} ((x + y) \cdot e^{y})

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} ln (x)

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