What is the integral from 0 to pi/3 of 4tan(x/2) ?

The integral from 0 to pi/3 of 4tan(x/2) is -8(1/2 ln(3)-ln(2))

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integral from 0 to pi/3 of 4tan(x/2)

\int_{0}^{\frac{π}{3}} 4 tan (\frac{x}{2}) d x

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

Decimal

1.15072 \dots

Solution steps

\int_{0}^{\frac{π}{3}} 4 tan (\frac{x}{2}) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{π}{3}} tan (\frac{x}{2}) d x

Rewrite using trig identities

= 4 \cdot \int_{0}^{\frac{π}{3}} \frac{sin ( \frac{x}{2} )}{cos ( \frac{x}{2} )} d x

Apply u-substitution

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

= 4 \cdot \int_{1}^{\frac{3}{2}} - \frac{2}{u} d u

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

= 4 (- \int_{\frac{3}{2}}^{1} - \frac{2}{u} d u)

Take the constant out:

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

= 4 (- (- 2 \cdot \int_{\frac{3}{2}}^{1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 (- (- 2 [ln ∣ u ∣]_{\frac{3}{2}}^{1}))

Simplify

= 8 [ln ∣ u ∣]_{\frac{3}{2}}^{1}

Compute the boundaries:

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

= 8 (- (\frac{1}{2} ln (3) - ln (2)))

Simplify

= - 8 (\frac{1}{2} ln (3) - ln (2))

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What is the integral from 0 to pi/3 of 4tan(x/2) ?
The integral from 0 to pi/3 of 4tan(x/2) is -8(1/2 ln(3)-ln(2))