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

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

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integral from 0 to 1/2 of tan(x/2)

\int_{0}^{\frac{1}{2}} tan (\frac{x}{2}) d x

- 2 ln (cos (\frac{1}{4}))

Decimal Notation

0.06316 \dots

Solution steps

\int_{0}^{\frac{1}{2}} tan (\frac{x}{2}) d x

Rewrite using trig identities

= \int_{0}^{\frac{1}{2}} \frac{sin ( \frac{x}{2} )}{cos ( \frac{x}{2} )} d x

Apply u-substitution

: \int_{1}^{c o s (\frac{1}{4})} - \frac{2}{u} d u

= \int_{1}^{c o s (\frac{1}{4})} - \frac{2}{u} d u

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

= - \int_{c o s (\frac{1}{4})}^{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_{c o s (\frac{1}{4})}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

Compute the boundaries:

- ln (cos (\frac{1}{4}))

= 2 (- ln (cos (\frac{1}{4})))

Simplify

= - 2 ln (cos (\frac{1}{4}))

x \to 0 lim (\frac{2 e ^{x} - 2}{x})

\frac{d y}{d t} = \frac{1}{10} ln (\frac{2000}{y}) y

\int csc^{2} (1) d x

f (x) = \frac{x ^{3} - 1}{2 x + 3}

(1 + 3 x^{2}) (x - x^{2})

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