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

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

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integral from 0 to pi/4 of tan(t)

\int_{0}^{\frac{π}{4}} tan (t) d t

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

Decimal

0.34657 \dots

Solution steps

\int_{0}^{\frac{π}{4}} tan (t) d t

Rewrite using trig identities

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{\frac{2}{2}}^{1})

Simplify

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

= [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

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

\int_{- 1}^{e - 2} \frac{1}{x + 2} d x

\int_{0}^{2} (x^{3} - 2 x + e^{x}) d x

\int_{0}^{2} (2) d x

\int_{0}^{1} 6 x e^{x} d x

\int_{0}^{40} q^{2} + 10 q - 90 d q

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