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

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

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

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

\frac{1}{3} - \frac{π}{6}

Decimal Notation

0.05375 \dots

Solution steps

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

Rewrite using trig identities

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

Expand

\frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} : \frac{1}{cos ^{2} ( x )} - 1

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

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

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

= \frac{1}{3} - \frac{π}{6}

\int cos^{3} (u) sin^{2} (u) d u

\int \frac{x ^{2} + 1}{x ^{2} - 1} d x

\frac{\partial}{\partial x} (- \frac{3}{x ^{2}})

\frac{d y}{d x}, y = x^{- 10 c o s (x)}

f (x) = \frac{x ^{2} - 1}{x ^{2} + x + 1}, (1, 0)

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