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

The integral from 0 to pi/3 of tan^3(x) is -ln(2)+3/2

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

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

- ln (2) + \frac{3}{2}

Decimal Notation

0.80685 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{3}} (- 1 + sec^{2} (x)) tan (x) d x

Apply u-substitution

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

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

Expand

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

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

\int_{1}^{2} \frac{1}{u} d u = ln (2)

\int_{1}^{2} u d u = \frac{3}{2}

= - ln (2) + \frac{3}{2}

\int cos^{3} (\frac{x}{19}) d x

\int (cos (x))^{3} d x

\frac{\partial}{\partial y} (arctan (x + 2 y))

\int \frac{5 - 5 x}{1 - x} d x

x^{^{''}} + 8 x^{^{'}} + 16 x = 0

What is the integral from 0 to pi/3 of tan^3(x) ?
The integral from 0 to pi/3 of tan^3(x) is -ln(2)+3/2