What is the integral from 0 to pi/(16) of tan(8x) ?

The integral from 0 to pi/(16) of tan(8x) is diverges

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integral from 0 to pi/(16) of tan(8x)

\int_{0}^{\frac{π}{16}} tan (8 x) d x

diverges

Solution steps

\int_{0}^{\frac{π}{16}} tan (8 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{16}} \frac{sin ( 8 x )}{cos ( 8 x )} d x

Apply u-substitution

: \int_{1}^{0} - \frac{1}{8 u} d u

= \int_{1}^{0} - \frac{1}{8 u} d u

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

= - \int_{0}^{1} - \frac{1}{8 u} d u

Take the constant out:

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

= - (- \frac{1}{8} \cdot \int_{0}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{8} [ln ∣ u ∣]_{0}^{1})

Simplify

= \frac{1}{8} [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int_{- 4}^{4} 4 - ∣ x ∣ d x

\frac{\partial}{\partial y} (cos (x + e^{y}) e^{y})

x \to 7 - lim (\frac{( x - 9 )}{x - 7})

t^{5} e^{3 t}

\int \frac{4}{x ^{2} - 8 x} d x

What is the integral from 0 to pi/(16) of tan(8x) ?
The integral from 0 to pi/(16) of tan(8x) is diverges