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

The integral from 0 to pi/6 of tan(3x) is diverges

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

\int_{0}^{\frac{π}{6}} tan (3 x) d x

diverges

Solution steps

\int_{0}^{\frac{π}{6}} tan (3 x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{0} - \frac{1}{3 u} d u

= \int_{1}^{0} - \frac{1}{3 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{1}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{3} \cdot \int_{0}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{3} [ln ∣ u ∣]_{0}^{1})

Simplify

= \frac{1}{3} [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

n = 1 \sum \infty \frac{ln ( n )}{n !}

\frac{d}{d x} (\frac{x}{1 + x})

\int 4 e^{- \frac{x}{4}} d x

f (x) = - 2 sin (x)

\frac{5 + 5 ^{2} + 4 x ^{2}}{2}

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