What is the integral of tan^3(y) ?

The integral of tan^3(y) is -ln|sec(y)|+(sec^2(y))/2+C

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integral of tan^3(y)

\int tan^{3} (y) d y

- ln ∣ sec (y) ∣ + \frac{sec ^{2} ( y )}{2} + C

Solution steps

\int tan^{3} (y) d y

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - ln ∣ u ∣ + \frac{u ^{2}}{2}

Substitute back

u = sec (y)

= - ln ∣ sec (y) ∣ + \frac{sec ^{2} ( y )}{2}

Add a constant to the solution

= - ln ∣ sec (y) ∣ + \frac{sec ^{2} ( y )}{2} + C

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