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

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

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

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

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

Decimal

0.86304 \dots

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

= 6 [ln ∣ u ∣]_{\frac{3}{2}}^{1}

Compute the boundaries:

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

= 6 (- (\frac{1}{2} ln (3) - ln (2)))

Simplify

= - 6 (\frac{1}{2} ln (3) - ln (2))

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