What is the integral from pi/8 to pi/6 of tan^2(2x) ?

The integral from pi/8 to pi/6 of tan^2(2x) is -pi/(24)+(sqrt(3)-1)/2

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integral from pi/8 to pi/6 of tan^2(2x)

\int_{\frac{π}{8}}^{\frac{π}{6}} tan^{2} (2 x) d x

- \frac{π}{24} + \frac{3 - 1}{2}

Decimal

0.23512 \dots

Solution steps

\int_{\frac{π}{8}}^{\frac{π}{6}} tan^{2} (2 x) d x

Rewrite using trig identities

= \int_{\frac{π}{8}}^{\frac{π}{6}} - 1 + sec^{2} (2 x) d x

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{π}{8}}^{\frac{π}{6}} 1 d x + \int_{\frac{π}{8}}^{\frac{π}{6}} sec^{2} (2 x) d x

\int_{\frac{π}{8}}^{\frac{π}{6}} 1 d x = \frac{π}{24}

\int_{\frac{π}{8}}^{\frac{π}{6}} sec^{2} (2 x) d x = \frac{3 - 1}{2}

= - \frac{π}{24} + \frac{3 - 1}{2}

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What is the integral from pi/8 to pi/6 of tan^2(2x) ?
The integral from pi/8 to pi/6 of tan^2(2x) is -pi/(24)+(sqrt(3)-1)/2