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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

= 2 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

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

Simplify

= ln (2)

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

\int ln (x^{\frac{1}{2}}) d x

\frac{s}{( s - 1 ) ^{2}}

\frac{\partial}{\partial x} (e^{e^{x}})

\frac{d}{d θ} (sin (θ) + cos (θ))

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