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

The integral from 0 to pi/6 of (sec(x)+tan(x))^2 is (-sqrt(3)pi+36)/(6sqrt(3))-2

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

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

\frac{- 3 π + 36}{6 3} - 2

Decimal

0.94050 \dots

Solution steps

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

Expand

(sec (x) + tan (x))^{2} : sec^{2} (x) + 2 sec (x) tan (x) + tan^{2} (x)

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

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

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

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

= \frac{1}{3} + 2 (\frac{2}{3} - 1) - \frac{π}{6} + \frac{1}{3}

Simplify

\frac{1}{3} + 2 (\frac{2}{3} - 1) - \frac{π}{6} + \frac{1}{3} : \frac{- 3 π + 36}{6 3} - 2

= \frac{- 3 π + 36}{6 3} - 2

What is the integral from 0 to pi/6 of (sec(x)+tan(x))^2 ?
The integral from 0 to pi/6 of (sec(x)+tan(x))^2 is (-sqrt(3)pi+36)/(6sqrt(3))-2