What is the integral from 0 to 1 of tan(pi/4 x) ?

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

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integral from 0 to 1 of tan(pi/4 x)

\int_{0}^{1} tan (\frac{π}{4} x) d x

\frac{2 ln ( 2 )}{π}

Decimal Notation

0.44127 \dots

Solution steps

\int_{0}^{1} tan (\frac{π}{4} x) d x

Rewrite using trig identities

= \int_{0}^{1} \frac{sin ( \frac{π x}{4} )}{cos ( \frac{π x}{4} )} d x

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

= \frac{4}{π} [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

= \frac{4}{π} \cdot \frac{1}{2} ln (2)

Simplify

= \frac{2 ln ( 2 )}{π}

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