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

f (x) = e^{x - 3}

e^{6 t}

y^{^{'}} = y - x

\int_{0}^{4} 16 - x^{3} d x

\frac{d y}{d x} - \frac{1}{x} \cdot y - 3 \cdot x^{2} = 0

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