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

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

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

\int_{- \frac{π}{6}}^{0} tan (x) d x

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

Decimal

- 0.14384 \dots

Solution steps

\int_{- \frac{π}{6}}^{0} tan (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

= \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}

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)

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