What is the integral of tan^3(10x) ?

The integral of tan^3(10x) is -1/10 ln|sec(10x)|+(sec^2(10x))/(20)+C

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integral of tan^3(10x)

\int tan^{3} (10 x) d x

- \frac{1}{10} ln ∣ sec (10 x) ∣ + \frac{sec ^{2} ( 10 x )}{20} + C

Solution steps

\int tan^{3} (10 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (10 x)) tan (10 x) d x

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{10 u} d u

= \int \frac{- 1 + u ^{2}}{10 u} d u

Expand

\frac{- 1 + u ^{2}}{10 u} : - \frac{1}{10 u} + \frac{u}{10}

= \int - \frac{1}{10 u} + \frac{u}{10} 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}{10 u} d u + \int \frac{u}{10} d u

\int \frac{1}{10 u} d u = \frac{1}{10} ln ∣ u ∣

\int \frac{u}{10} d u = \frac{u ^{2}}{20}

= - \frac{1}{10} ln ∣ u ∣ + \frac{u ^{2}}{20}

Substitute back

u = sec (10 x)

= - \frac{1}{10} ln ∣ sec (10 x) ∣ + \frac{sec ^{2} ( 10 x )}{20}

Add a constant to the solution

= - \frac{1}{10} ln ∣ sec (10 x) ∣ + \frac{sec ^{2} ( 10 x )}{20} + C

\int \frac{5 x - 3}{x ^{2} - 2 x - 3} d x

\frac{\partial}{\partial x} (\frac{x + y}{x - y})

\frac{d}{d x} (sin (cos (4 x)))

\frac{1}{5 x ^{\frac{4}{5}}}

f (x) = \frac{x ^{3} + 1}{x ^{3} - 1}

What is the integral of tan^3(10x) ?
The integral of tan^3(10x) is -1/10 ln|sec(10x)|+(sec^2(10x))/(20)+C