What is the integral from pi/6 to pi/3 of tan(x) ?

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

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integral from pi/6 to pi/3 of tan(x)

\int_{\frac{π}{6}}^{\frac{π}{3}} tan (x) d x

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

Solution steps

\int_{\frac{π}{6}}^{\frac{π}{3}} tan (x) d x

Rewrite using trig identities

= \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{sin ( x )}{cos ( x )} d x

Apply u-substitution

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

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

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

= [ln ∣ u ∣]_{\frac{1}{2}}^{\frac{3}{2}}

Compute the boundaries:

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

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

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\int_{0}^{0.1} \frac{1}{1 + x ^{4}} d x

\int_{0}^{1} (2 + e^{x})^{2} d x

\int_{- 3}^{1} (2 x + 1) (3 + x) (1 - x) d x

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