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

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

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

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

- \frac{π}{6} + \frac{1}{3}

Decimal

0.05375 \dots

Solution steps

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

Rewrite using trig identities

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

\int_{0}^{\frac{π}{6}} 1 d x = \frac{π}{6}

\int_{0}^{\frac{π}{6}} sec^{2} (x) d x = \frac{1}{3}

= - \frac{π}{6} + \frac{1}{3}

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