What is the integral of tan^3(θ)sec(θ) ?

The integral of tan^3(θ)sec(θ) is -sec(θ)+(sec^3(θ))/3+C

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integral of tan^3(θ)sec(θ)

\int tan^{3} (θ) sec (θ) d θ

- sec (θ) + \frac{sec ^{3} ( θ )}{3} + C

Solution steps

\int tan^{3} (θ) sec (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

: \int - 1 + u^{2} d u

= \int - 1 + u^{2} d u

Apply the Sum Rule:

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

= - \int 1 d u + \int u^{2} d u

\int 1 d u = u

\int u^{2} d u = \frac{u ^{3}}{3}

= - u + \frac{u ^{3}}{3}

Substitute back

u = sec (θ)

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

Add a constant to the solution

= - sec (θ) + \frac{sec ^{3} ( θ )}{3} + C

\frac{d y}{d t} + 2 y = t e^{- 2 t}

\frac{\partial}{\partial x} (- 3 e^{x} ln (y))

\int \frac{x ^{3}}{x ^{2} - 4} d x

x \to 1 + lim (\frac{( 2 )}{x - 1})

\int_{- 2}^{3} (x^{2} - 4) d x

What is the integral of tan^3(θ)sec(θ) ?
The integral of tan^3(θ)sec(θ) is -sec(θ)+(sec^3(θ))/3+C