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

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

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

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

4 ln (2)

Decimal

2.77258 \dots

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{\frac{1}{2}} - \frac{4}{u} d u

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

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

= 4 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

ln (2)

= 4 ln (2)

\int_{x}^{x + 4} \frac{1}{t ^{2} + 2} d t

\int_{- 1}^{5} ∣ x - 3 ∣ d x

\int_{1}^{4} 7 x^{- \frac{3}{2}} d x

\int_{0}^{3} (4 x) d x

\int_{0}^{1} x \cdot 1 + 4 x^{2} d x

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