What is the integral of tan^3(5x) ?

The integral of tan^3(5x) is -1/5 ln|sec(5x)|+(sec^2(5x))/(10)+C

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integral of tan^3(5x)

\int tan^{3} (5 x) d x

- \frac{1}{5} ln ∣ sec (5 x) ∣ + \frac{sec ^{2} ( 5 x )}{10} + C

Solution steps

\int tan^{3} (5 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (5 x)) tan (5 x) d x

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{5 u} d u

= \int \frac{- 1 + u ^{2}}{5 u} d u

Expand

\frac{- 1 + u ^{2}}{5 u} : - \frac{1}{5 u} + \frac{u}{5}

= \int - \frac{1}{5 u} + \frac{u}{5} 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}{5 u} d u + \int \frac{u}{5} d u

\int \frac{1}{5 u} d u = \frac{1}{5} ln ∣ u ∣

\int \frac{u}{5} d u = \frac{u ^{2}}{10}

= - \frac{1}{5} ln ∣ u ∣ + \frac{u ^{2}}{10}

Substitute back

u = sec (5 x)

= - \frac{1}{5} ln ∣ sec (5 x) ∣ + \frac{sec ^{2} ( 5 x )}{10}

Add a constant to the solution

= - \frac{1}{5} ln ∣ sec (5 x) ∣ + \frac{sec ^{2} ( 5 x )}{10} + C

y = (ln ∣ x + 8 ∣)^{7}

y^{^{'}} = \frac{( 3 y ^{2} - 9 )}{2 x y}

\int (sin (x))^{3} (cos (x))^{3} d x

e^{a r c s i n (x)}

\frac{d y}{d x} = - \frac{1}{2 y}

What is the integral of tan^3(5x) ?
The integral of tan^3(5x) is -1/5 ln|sec(5x)|+(sec^2(5x))/(10)+C