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

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

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

\int_{0}^{π} 4 tan (\frac{x}{4}) d x

8 ln (2)

Decimal

5.54517 \dots

Solution steps

\int_{0}^{π} 4 tan (\frac{x}{4}) d x

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

4 (- (- 4 [ln ∣ u ∣]_{\frac{2}{2}}^{1})) : 16 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= 16 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

= 16 \cdot \frac{1}{2} ln (2)

Simplify

= 8 ln (2)

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