What is the integral of tan^3(11pix) ?

The integral of tan^3(11pix) is 1/(11pi)(-ln|sec(11pix)|+(sec^2(11pix))/2)+C

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integral of tan^3(11pix)

\int tan^{3} (11 π x) d x

\frac{1}{11 π} (- ln ∣ sec (11 π x) ∣ + \frac{sec ^{2} ( 11 π x )}{2}) + C

Solution steps

\int tan^{3} (11 π x) d x

Apply u-substitution

: \int tan^{3} (u) \frac{1}{11 π} d u

= \int tan^{3} (u) \frac{1}{11 π} d u

Take the constant out:

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

= \frac{1}{11 π} \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= \frac{1}{11 π} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

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

= \frac{1}{11 π} \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

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

= \frac{1}{11 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}{11 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}{11 π} (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= \frac{1}{11 π} (- ln ∣ sec (11 π x) ∣ + \frac{sec ^{2} ( 11 π x )}{2})

Add a constant to the solution

= \frac{1}{11 π} (- ln ∣ sec (11 π x) ∣ + \frac{sec ^{2} ( 11 π x )}{2}) + C

\frac{d}{d x} (\frac{1}{4} sin (2 x))

\int (x^{\frac{1}{2}}) d x

\frac{d y}{d x} + 5 y = 20

y^{^{''}} + y = cot^{2} (x)

\frac{( s + 4 )}{( s ^{2} - 2 s + 4 )}

What is the integral of tan^3(11pix) ?
The integral of tan^3(11pix) is 1/(11pi)(-ln|sec(11pix)|+(sec^2(11pix))/2)+C