What is the integral of 12tan^3(x) ?

The integral of 12tan^3(x) is 12(-ln|sec(x)|+(sec^2(x))/2)+C

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integral of 12tan^3(x)

\int 12 tan^{3} (x) d x

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

Solution steps

\int 12 tan^{3} (x) d x

Take the constant out:

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

= 12 \cdot \int tan^{3} (x) d x

Rewrite using trig identities

= 12 \cdot \int (- 1 + sec^{2} (x)) tan (x) d x

Apply u-substitution

= 12 \cdot \int \frac{- 1 + u ^{2}}{u} d u

Expand

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

= 12 \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

= 12 (- \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}

= 12 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = sec (x)

= 12 (- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2})

Add a constant to the solution

= 12 (- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2}) + C

\int_{0}^{1} \frac{42}{4 y - 1} d y

y^{^{''}} = 6 x + 2, y (1) = - 1, y^{^{'}} (1) = 7

\frac{\partial}{\partial z} (arctan (x^{2} y z^{2}))

\frac{\partial}{\partial x} ((2 x + 3 y)^{2})

x \to 0 lim (\frac{1}{x - 3})

What is the integral of 12tan^3(x) ?
The integral of 12tan^3(x) is 12(-ln|sec(x)|+(sec^2(x))/2)+C