What is the integral from pi/6 to pi/4 of tan(x) ?

The integral from pi/6 to pi/4 of tan(x) is (ln(3)-ln(2))/2

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integral from pi/6 to pi/4 of tan(x)

\int_{\frac{π}{6}}^{\frac{π}{4}} tan (x) d x

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

Decimal

0.20273 \dots

Solution steps

\int_{\frac{π}{6}}^{\frac{π}{4}} tan (x) d x

Rewrite using trig identities

= \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{sin ( x )}{cos ( x )} d x

Apply u-substitution

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

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

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

Compute the boundaries:

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

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

a re a y^{2} - 4 x = 4, 4 x - y = 16

\frac{d}{d x} ((\frac{- sin ( x )}{1 - cos ( x )})^{2})

\int x (x - 1) d x

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

\frac{\partial}{\partial x} (\frac{m \cdot x \cdot T}{A})

What is the integral from pi/6 to pi/4 of tan(x) ?
The integral from pi/6 to pi/4 of tan(x) is (ln(3)-ln(2))/2