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

The integral from 0 to pi of tan(x/6) is -6(1/2 ln(3)-ln(2))

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

\int_{0}^{π} tan (\frac{x}{6}) d x

- 6 (\frac{1}{2} ln (3) - ln (2))

Decimal

0.86304 \dots

Solution steps

\int_{0}^{π} tan (\frac{x}{6}) d x

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{\frac{3}{2}} - \frac{6}{u} d u

= \int_{1}^{\frac{3}{2}} - \frac{6}{u} d u

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

= - \int_{\frac{3}{2}}^{1} - \frac{6}{u} d u

Take the constant out:

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

= - (- 6 \cdot \int_{\frac{3}{2}}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- 6 [ln ∣ u ∣]_{\frac{3}{2}}^{1})

Simplify

= 6 [ln ∣ u ∣]_{\frac{3}{2}}^{1}

Compute the boundaries:

- (\frac{1}{2} ln (3) - ln (2))

= 6 (- (\frac{1}{2} ln (3) - ln (2)))

Simplify

= - 6 (\frac{1}{2} ln (3) - ln (2))

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What is the integral from 0 to pi of tan(x/6) ?
The integral from 0 to pi of tan(x/6) is -6(1/2 ln(3)-ln(2))