What is the integral from 0 to pi/8 of tan(4x) ?

The integral from 0 to pi/8 of tan(4x) is diverges

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integral from 0 to pi/8 of tan(4x)

\int_{0}^{\frac{π}{8}} tan (4 x) d x

diverges

Solution steps

\int_{0}^{\frac{π}{8}} tan (4 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{8}} \frac{sin ( 4 x )}{cos ( 4 x )} d x

Apply u-substitution

: \int_{1}^{0} - \frac{1}{4 u} d u

= \int_{1}^{0} - \frac{1}{4 u} d u

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

= - \int_{0}^{1} - \frac{1}{4 u} d u

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{0}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{4} [ln ∣ u ∣]_{0}^{1})

Simplify

= \frac{1}{4} [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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What is the integral from 0 to pi/8 of tan(4x) ?
The integral from 0 to pi/8 of tan(4x) is diverges