What is the integral of (tan^3(x)) ?

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

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integral of (tan^3(x))

\int (tan^{3} (x)) d x

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

Solution steps

\int tan^{3} (x) d x

Rewrite using trig identities

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

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 (x)

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

Add a constant to the solution

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

y = x^{2}, y = x + 2, 0

y = \frac{x}{x ^{2} - 4 x + 5}, y = \frac{3}{x ^{2} - 4 x + 5}, x = 0

y = \frac{x ^{3} + 1}{6 x}, x = 1

y^{^{'}} + 4 y = e^{- x}, y (0) = 10

\int_{0}^{6} \frac{1}{∣ x - 3 ∣} d x

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