What is the integral of (tan^3(x))/2 ?

The integral of (tan^3(x))/2 is 1/2 (-ln|sec(x)|+(sec^2(x))/2)+C

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integral of (tan^3(x))/2

\int \frac{tan ^{3} ( x )}{2} d x

\frac{1}{2} (- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2}) + C

Solution steps

\int \frac{tan ^{3} ( x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int tan^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \frac{1}{2} \cdot \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

= \frac{1}{2} (- \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}

= \frac{1}{2} (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = sec (x)

= \frac{1}{2} (- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2})

Add a constant to the solution

= \frac{1}{2} (- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2}) + C

\frac{\partial}{\partial x} (cos (- 5 x - 6 y))

e^{x} x

\int e^{7 x} cos (9 x) d x

t^{2} + t + 1

x \to 0 lim (\frac{cos ( 2 x ) - 1}{x})

What is the integral of (tan^3(x))/2 ?
The integral of (tan^3(x))/2 is 1/2 (-ln|sec(x)|+(sec^2(x))/2)+C