What is the integral of tan^3(7pix) ?

The integral of tan^3(7pix) is 1/(7pi)(-ln|sec(7pix)|+(sec^2(7pix))/2)+C

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integral of tan^3(7pix)

\int tan^{3} (7 π x) d x

\frac{1}{7 π} (- ln ∣ sec (7 π x) ∣ + \frac{sec ^{2} ( 7 π x )}{2}) + C

Solution steps

\int tan^{3} (7 π x) d x

Apply u-substitution

: \int tan^{3} (u) \frac{1}{7 π} d u

= \int tan^{3} (u) \frac{1}{7 π} d u

Take the constant out:

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

= \frac{1}{7 π} \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= \frac{1}{7 π} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

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

= \frac{1}{7 π} \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

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

= \frac{1}{7 π} \cdot \int - \frac{1}{v} + v d v

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}{7 π} (- \int \frac{1}{v} d v + \int v d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int v d v = \frac{v ^{2}}{2}

= \frac{1}{7 π} (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= \frac{1}{7 π} (- ln ∣ sec (7 π x) ∣ + \frac{sec ^{2} ( 7 π x )}{2})

Add a constant to the solution

= \frac{1}{7 π} (- ln ∣ sec (7 π x) ∣ + \frac{sec ^{2} ( 7 π x )}{2}) + C

\frac{0.5 s}{5 s ^{2} + 10 s + 49}

n = 1 \sum \infty \frac{1}{e ^{n}}

\int \frac{x ^{2} - 8}{8 x} d x

\frac{d}{d x} (ln (arctan (9 x^{4})))

f (x) = ln (x)

What is the integral of tan^3(7pix) ?
The integral of tan^3(7pix) is 1/(7pi)(-ln|sec(7pix)|+(sec^2(7pix))/2)+C