What is the integral of tan^3(7x+8) ?

The integral of tan^3(7x+8) is 1/7 (-ln|sec(7x+8)|+(sec^2(7x+8))/2)+C

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integral of tan^3(7x+8)

\int tan^{3} (7 x + 8) d x

\frac{1}{7} (- ln ∣ sec (7 x + 8) ∣ + \frac{sec ^{2} ( 7 x + 8 )}{2}) + C

Solution steps

\int tan^{3} (7 x + 8) d x

Apply u-substitution

: \int tan^{3} (u) \frac{1}{7} d u

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

Substitute back

= \frac{1}{7} (- ln ∣ sec (7 x + 8) ∣ + \frac{sec ^{2} ( 7 x + 8 )}{2})

Add a constant to the solution

= \frac{1}{7} (- ln ∣ sec (7 x + 8) ∣ + \frac{sec ^{2} ( 7 x + 8 )}{2}) + C

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What is the integral of tan^3(7x+8) ?
The integral of tan^3(7x+8) is 1/7 (-ln|sec(7x+8)|+(sec^2(7x+8))/2)+C