What is the integral from 0 to pi/6 of tan^3(2x) ?

The integral from 0 to pi/6 of tan^3(2x) is -1/2 ln(2)+3/4

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integral from 0 to pi/6 of tan^3(2x)

\int_{0}^{\frac{π}{6}} tan^{3} (2 x) d x

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

Decimal

0.40342 \dots

Solution steps

\int_{0}^{\frac{π}{6}} tan^{3} (2 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{6}} (- 1 + sec^{2} (2 x)) tan (2 x) d x

Apply u-substitution

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

Expand

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

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

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

\int_{1}^{2} \frac{u}{2} d u = \frac{3}{4}

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

What is the integral from 0 to pi/6 of tan^3(2x) ?
The integral from 0 to pi/6 of tan^3(2x) is -1/2 ln(2)+3/4