What is the integral of sec^3(θ) ?

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

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integral of sec^3(θ)

\int sec^{3} (θ) d θ

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

Solution steps

\int sec^{3} (θ) d θ

Apply Integral Reduction

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{3} - 2 x^{2} + y^{3} - 2 y)

\frac{d}{d x} ((sin^{2} (x)) d x)

\int \frac{x ^{2} + 5 x ^{3} - 7}{x ^{2}} d x

\int \frac{9 x}{1 + x ^{4}} d x

x \to 2 - lim (x + 5)

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