What is the integral from 0 to pi of tan^2(x) ?

The integral from 0 to pi of tan^2(x) is diverges

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integral from 0 to pi of tan^2(x)

\int_{0}^{π} tan^{2} (x) d x

diverges

Solution steps

\int_{0}^{π} tan^{2} (x) d x

Take the denominator(s) of

tan^{2} (x)

and compare to zero:

x = \frac{π}{2}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{π}{2}} tan^{2} (x) d x + \int_{\frac{π}{2}}^{π} tan^{2} (x) d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (\frac{x ^{2} - x + 2}{x})

y = 2 x^{2}, y = 16 x - 32, y = 0

\frac{d y}{d t} = \frac{y + 1}{t + 1}

\int \frac{1}{1 - x ^{2}} d x

\frac{\partial}{\partial x} (6 - arctan (7 x^{2} + 2 y^{2}))

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