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

The integral of tan^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 tan^2(θ)sec(θ)

\int tan^{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 tan^{2} (θ) sec (θ) d θ

Rewrite using trig identities

= \int (- 1 + sec^{2} (θ)) sec (θ) d θ

Expand

(- 1 + sec^{2} (θ)) sec (θ) : - sec (θ) + sec^{3} (θ)

= \int - sec (θ) + sec^{3} (θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int sec (θ) d θ + \int sec^{3} (θ) d θ

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

\int sec^{3} (θ) d θ = \frac{1}{2} sec (θ) tan (θ) + \frac{1}{2} ln ∣ tan (θ) + sec (θ) ∣

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

Simplify

- 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) ∣

= \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 - 4 lo g_{7} (x) d x

\int x^{2} - 7 x d x

\int (3 sin (x)) d x

f (x) = \frac{x + 7}{x}

n = 1 \sum \infty \frac{1}{3 ^{n - 3}}

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