What is the integral of tan^3(θ) ?

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

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integral of tan^3(θ)

\int tan^{3} (θ) d θ

- ln ∣ sec (θ) ∣ + \frac{sec ^{2} ( θ )}{2} + C

Solution steps

\int tan^{3} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{- 1 + u ^{2}}{u} : - \frac{1}{u} + u

= \int - \frac{1}{u} + u 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 \frac{1}{u} d u + \int u d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - ln ∣ u ∣ + \frac{u ^{2}}{2}

Substitute back

u = sec (θ)

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

Add a constant to the solution

= - ln ∣ sec (θ) ∣ + \frac{sec ^{2} ( θ )}{2} + C

x y^{^{''}} + 2 y^{^{'}} = 8 x^{2}

\frac{\partial}{\partial y} (x e^{x y^{2}})

x \to - 1 - lim (- 4 x - 4 - 8)

y^{^{'}} + y = \frac{1 - x}{2 x ^{2}}

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