What is the integral of (tan(x/(10)))^5 ?

The integral of (tan(x/(10)))^5 is 10(ln|sec(x/(10))|-sec^2(x/(10))+(sec^4(x/(10)))/4)+C

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integral of (tan(x/(10)))^5

\int (tan (\frac{x}{10}))^{5} d x

10 (ln sec (\frac{x}{10}) - sec^{2} (\frac{x}{10}) + \frac{sec ^{4} ( \frac{x}{10} )}{4}) + C

Solution steps

\int (tan (\frac{x}{10}))^{5} d x

Simplify

= \int tan^{5} (\frac{x}{10}) d x

Apply u-substitution

: \int tan^{5} (u) \cdot 10 d u

= \int tan^{5} (u) \cdot 10 d u

Take the constant out:

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

= 10 \cdot \int tan^{5} (u) d u

Rewrite using trig identities

= 10 \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) d u

Apply u-substitution

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

= 10 \cdot \int \frac{( - 1 + v ^{2} ) ^{2}}{v} d v

Expand

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

= 10 \cdot \int \frac{1}{v} - 2 v + v^{3} 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

= 10 (\int \frac{1}{v} d v - \int 2 v d v + \int v^{3} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

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

= 10 (ln ∣ v ∣ - v^{2} + \frac{v ^{4}}{4})

Substitute back

= 10 (ln sec (\frac{x}{10}) - sec^{2} (\frac{x}{10}) + \frac{sec ^{4} ( \frac{x}{10} )}{4})

Add a constant to the solution

= 10 (ln sec (\frac{x}{10}) - sec^{2} (\frac{x}{10}) + \frac{sec ^{4} ( \frac{x}{10} )}{4}) + C

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

x^{2} + y^{2} = 25, (- 4, - 3)

\int 2 x sin (x^{2}) d x

y^{^{'}} + \frac{y ^{2}}{x} = - \frac{1}{x}

\int \frac{x ^{2} - 1}{x ^{2} + x} d x

What is the integral of (tan(x/(10)))^5 ?
The integral of (tan(x/(10)))^5 is 10(ln|sec(x/(10))|-sec^2(x/(10))+(sec^4(x/(10)))/4)+C