What is the integral of sec^2(θ)sec(θ) ?

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

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integral of sec^2(θ)sec(θ)

\int sec^{2} (θ) sec (θ) d θ

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

Solution steps

\int sec^{2} (θ) sec (θ) d θ

Simplify

sec^{2} (θ) sec (θ) : sec^{3} (θ)

= \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

\int \frac{( 16 x ^{3} )}{x ^{4} + 5} d x

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