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

The integral from 0 to pi/2 of tan(t) is diverges

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

\int_{0}^{\frac{π}{2}} tan (t) d t

diverges

Solution steps

\int_{0}^{\frac{π}{2}} tan (t) d t

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{0} - \frac{1}{u} d u

= \int_{1}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{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_{0}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{0}^{1})

Simplify

= [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{π}{2}} sin^{3} (y) d y

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

\int_{0}^{5} t e^{- 0.4 t} d t

\int_{0}^{2} - x^{2} + 2 x + 1 d x

\int_{0}^{1} x + x d x

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