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

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

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integral of (tan(x))^5

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{sin ^{5} ( x )}{cos ^{5} ( x )} d x

Rewrite using trig identities

= \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

\frac{d}{d x} (ln (1 + x^{3}))

y = x^{3} - 2 x + 1, (0, 1)

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\int_{1}^{2} \frac{15}{x ^{3} + 4 x} d x

n = 1 \sum \infty n^{50} e^{- n}

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