What is the integral of 16tan^3(x) ?

The integral of 16tan^3(x) is 16(-ln|sec(x)|+(sec^2(x))/2)+C

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integral of 16tan^3(x)

\int 16 tan^{3} (x) d x

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

Solution steps

\int 16 tan^{3} (x) d x

Take the constant out:

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

= 16 \cdot \int tan^{3} (x) d x

Rewrite using trig identities

= 16 \cdot \int (- 1 + sec^{2} (x)) tan (x) d x

Apply u-substitution

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

= 16 \cdot \int \frac{- 1 + u ^{2}}{u} d u

Expand

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

= 16 \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

= 16 (- \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}

= 16 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = sec (x)

= 16 (- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2})

Add a constant to the solution

= 16 (- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2}) + C

g (w) = 10 w

\frac{\partial}{\partial x} (x^{4} y)

\int t \cdot sin (2 t) d t

(8 (5 x)^{4} + 5)^{3}

\frac{d}{d x} (\frac{1}{x ^{4}} - \frac{3}{x ^{2}} + x^{3})

What is the integral of 16tan^3(x) ?
The integral of 16tan^3(x) is 16(-ln|sec(x)|+(sec^2(x))/2)+C