What is the integral from 0 to pi/(28) of tan(14x) ?

The integral from 0 to pi/(28) of tan(14x) is diverges

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integral from 0 to pi/(28) of tan(14x)

\int_{0}^{\frac{π}{28}} tan (14 x) d x

diverges

Solution steps

\int_{0}^{\frac{π}{28}} tan (14 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{28}} \frac{sin ( 14 x )}{cos ( 14 x )} d x

Apply u-substitution

: \int_{1}^{0} - \frac{1}{14 u} d u

= \int_{1}^{0} - \frac{1}{14 u} d u

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

= - \int_{0}^{1} - \frac{1}{14 u} d u

Take the constant out:

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

= - (- \frac{1}{14} \cdot \int_{0}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{14} [ln ∣ u ∣]_{0}^{1})

Simplify

= \frac{1}{14} [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (\frac{3 x}{5 y})

\frac{m x}{b x + c}

\int (x + 5) 8 - x d x

\frac{\partial}{\partial x} (2 + 3 x y)

n = 0 \sum \infty (4 x^{2})^{n}

What is the integral from 0 to pi/(28) of tan(14x) ?
The integral from 0 to pi/(28) of tan(14x) is diverges