What is the integral from 0 to pi/(24) of tan(8x) ?

The integral from 0 to pi/(24) of tan(8x) is 1/8 ln(2)

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integral from 0 to pi/(24) of tan(8x)

\int_{0}^{\frac{π}{24}} tan (8 x) d x

\frac{1}{8} ln (2)

Decimal Notation

0.08664 \dots

Solution steps

\int_{0}^{\frac{π}{24}} tan (8 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{24}} \frac{sin ( 8 x )}{cos ( 8 x )} d x

Apply u-substitution

: \int_{1}^{\frac{1}{2}} - \frac{1}{8 u} d u

= \int_{1}^{\frac{1}{2}} - \frac{1}{8 u} d u

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

= - \int_{\frac{1}{2}}^{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_{\frac{1}{2}}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{8} [ln ∣ u ∣]_{\frac{1}{2}}^{1})

Simplify

= \frac{1}{8} [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

ln (2)

= \frac{1}{8} ln (2)

e^{1 + x^{2}} + 3 x^{4}

\frac{d y}{d x} y = sin (- 1 (6 x + 1))

\int 10 x 3 x^{2} - 4 d x

\frac{\partial}{\partial y} (x^{2} + 2 y^{2} - x^{2} y)

x \to \frac{7 π}{2} + lim (sec (x))

What is the integral from 0 to pi/(24) of tan(8x) ?
The integral from 0 to pi/(24) of tan(8x) is 1/8 ln(2)