What is the integral from 0 to x/4 of tan^3(x) ?

The integral from 0 to x/4 of tan^3(x) is 1/2 tan^2(x/4)-ln|sec(x/4)|

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integral from 0 to x/4 of tan^3(x)

\int_{0}^{\frac{x}{4}} tan^{3} (x) d x

\frac{1}{2} tan^{2} (\frac{x}{4}) - ln sec (\frac{x}{4})

Solution steps

\int_{0}^{\frac{x}{4}} tan^{3} (x) d x

Rewrite using trig identities

= \int_{0}^{\frac{x}{4}} (- 1 + sec^{2} (x)) tan (x) d x

Apply u-substitution

: \int_{1}^{s e c (\frac{x}{4})} \frac{- 1 + u ^{2}}{u} d u

= \int_{1}^{s e c (\frac{x}{4})} \frac{- 1 + u ^{2}}{u} d u

Expand

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

= \int_{1}^{s e c (\frac{x}{4})} - \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_{1}^{s e c (\frac{x}{4})} \frac{1}{u} d u + \int_{1}^{s e c (\frac{x}{4})} u d u

\int_{1}^{s e c (\frac{x}{4})} \frac{1}{u} d u = ln sec (\frac{x}{4})

\int_{1}^{s e c (\frac{x}{4})} u d u = \frac{sec ^{2} ( \frac{x}{4} ) - 1}{2}

= - ln sec (\frac{x}{4}) + \frac{sec ^{2} ( \frac{x}{4} ) - 1}{2}

Simplify

- ln sec (\frac{x}{4}) + \frac{sec ^{2} ( \frac{x}{4} ) - 1}{2} : \frac{1}{2} tan^{2} (\frac{x}{4}) - ln sec (\frac{x}{4})

= \frac{1}{2} tan^{2} (\frac{x}{4}) - ln sec (\frac{x}{4})

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What is the integral from 0 to x/4 of tan^3(x) ?
The integral from 0 to x/4 of tan^3(x) is 1/2 tan^2(x/4)-ln|sec(x/4)|