What is the integral from 1 to pi/3 of tan(t) ?

The integral from 1 to pi/3 of tan(t) is ln(2cos(1))

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integral from 1 to pi/3 of tan(t)

\int_{1}^{\frac{π}{3}} tan (t) d t

ln (2 cos (1))

Decimal

0.07752 \dots

Solution steps

\int_{1}^{\frac{π}{3}} tan (t) d t

Rewrite using trig identities

= \int_{1}^{\frac{π}{3}} \frac{sin ( t )}{cos ( t )} d t

Apply u-substitution

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

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

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

Compute the boundaries:

ln (cos (1)) + ln (2)

= ln (cos (1)) + ln (2)

Simplify

= ln (2 cos (1))

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What is the integral from 1 to pi/3 of tan(t) ?
The integral from 1 to pi/3 of tan(t) is ln(2cos(1))