What is the integral from 0 to pi/(12) of tan(6x) ?

The integral from 0 to pi/(12) of tan(6x) is diverges

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integral from 0 to pi/(12) of tan(6x)

\int_{0}^{\frac{π}{12}} tan (6 x) d x

diverges

Solution steps

\int_{0}^{\frac{π}{12}} tan (6 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{12}} \frac{sin ( 6 x )}{cos ( 6 x )} d x

Apply u-substitution

: \int_{1}^{0} - \frac{1}{6 u} d u

= \int_{1}^{0} - \frac{1}{6 u} d u

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

= - \int_{0}^{1} - \frac{1}{6 u} d u

Take the constant out:

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

= - (- \frac{1}{6} \cdot \int_{0}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{6} [ln ∣ u ∣]_{0}^{1})

Simplify

= \frac{1}{6} [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\frac{x + 6}{x ^{3} + x - 7}

\int x (x + 4)^{6} d x

\int_{0}^{1} cos^{2} (x) d x

\frac{d}{d x} (ln (6 x + 1))

\frac{\partial}{\partial x} (yz - ln (x + z))

What is the integral from 0 to pi/(12) of tan(6x) ?
The integral from 0 to pi/(12) of tan(6x) is diverges