What is the integral from pi/6 to pi/3 of tan^3(x) ?

The integral from pi/6 to pi/3 of tan^3(x) is -1/2 ln(3)+4/3

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integral from pi/6 to pi/3 of tan^3(x)

\int_{\frac{π}{6}}^{\frac{π}{3}} tan^{3} (x) d x

- \frac{1}{2} ln (3) + \frac{4}{3}

Decimal

0.78402 \dots

Solution steps

\int_{\frac{π}{6}}^{\frac{π}{3}} tan^{3} (x) d x

Rewrite using trig identities

= \int_{\frac{π}{6}}^{\frac{π}{3}} (- 1 + sec^{2} (x)) tan (x) d x

Apply u-substitution

: \int_{\frac{2 3}{3}}^{2} \frac{- 1 + u ^{2}}{u} d u

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

Expand

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

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

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

\int_{\frac{2 3}{3}}^{2} u d u = \frac{4}{3}

= - \frac{1}{2} ln (3) + \frac{4}{3}

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What is the integral from pi/6 to pi/3 of tan^3(x) ?
The integral from pi/6 to pi/3 of tan^3(x) is -1/2 ln(3)+4/3