What is the integral from 0 to pi/(16) of 24tan(4x) ?

The integral from 0 to pi/(16) of 24tan(4x) is 3ln(2)

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integral from 0 to pi/(16) of 24tan(4x)

\int_{0}^{\frac{π}{16}} 24 tan (4 x) d x

3 ln (2)

Decimal Notation

2.07944 \dots

Solution steps

\int_{0}^{\frac{π}{16}} 24 tan (4 x) d x

Take the constant out:

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

= 24 \cdot \int_{0}^{\frac{π}{16}} tan (4 x) d x

Rewrite using trig identities

= 24 \cdot \int_{0}^{\frac{π}{16}} \frac{sin ( 4 x )}{cos ( 4 x )} d x

Apply u-substitution

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

= 24 \cdot \int_{1}^{\frac{2}{2}} - \frac{1}{4 u} d u

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

= 24 (- \int_{\frac{2}{2}}^{1} - \frac{1}{4 u} d u)

Take the constant out:

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

= 24 (- (- \frac{1}{4} \cdot \int_{\frac{2}{2}}^{1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 24 (- (- \frac{1}{4} [ln ∣ u ∣]_{\frac{2}{2}}^{1}))

Simplify

24 (- (- \frac{1}{4} [ln ∣ u ∣]_{\frac{2}{2}}^{1})) : 6 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= 6 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

= 6 \cdot \frac{1}{2} ln (2)

Simplify

= 3 ln (2)

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What is the integral from 0 to pi/(16) of 24tan(4x) ?
The integral from 0 to pi/(16) of 24tan(4x) is 3ln(2)