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

The integral of tan^5(x) is ln|sec(x)|-sec^2(x)+(sec^4(x))/4+C

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integral of tan^5(x)

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

ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{( - 1 + u ^{2} ) ^{2}}{u} : \frac{1}{u} - 2 u + u^{3}

= \int \frac{1}{u} - 2 u + u^{3} 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 2 u d u + \int u^{3} d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int u^{3} d u = \frac{u ^{4}}{4}

= ln ∣ u ∣ - u^{2} + \frac{u ^{4}}{4}

Substitute back

u = sec (x)

= ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4}

Add a constant to the solution

= ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4} + C

(x + y) e^{\frac{x}{y}} d x + (x - \frac{x ^{2}}{y}) e^{\frac{x}{y}} d y = 0

\int (5 x^{14} - 2 x^{5}) d x

\int (y^{2} - 1)^{2} d y

\frac{d}{d x} (x^{2} + a^{2})

\int e^{3 x - 4} d x

What is the integral of tan^5(x) ?
The integral of tan^5(x) is ln|sec(x)|-sec^2(x)+(sec^4(x))/4+C