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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{0}^{\frac{π}{6}} tan (x) d x

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

Decimal

0.14384 \dots

Solution steps

\int_{0}^{\frac{π}{6}} tan (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{\frac{3}{2}}^{1})

Simplify

= [ln ∣ u ∣]_{\frac{3}{2}}^{1}

Compute the boundaries:

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

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

Simplify

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

\int_{0}^{5} (u^{2} + u + 1) d u

\int_{- π}^{0} - 3 cos^{2} (x) sin (x) d x

\int_{- 4}^{6} ∣ x ∣ d x

\int_{- 2}^{- 1} (x^{2} + \frac{1}{x ^{2}}) d x

\int_{1}^{4} x^{2} ln (x) d x

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