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

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

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integral from 0 to 1 of tan(x)

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

- ln (cos (1))

Decimal

0.61562 \dots

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{c o s (1)}^{1})

Simplify

= [ln ∣ u ∣]_{c o s (1)}^{1}

Compute the boundaries:

- ln (cos (1))

= - ln (cos (1))

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