What is the integral from 0 to 4 of sec(x)tan(x) ?

The integral from 0 to 4 of sec(x)tan(x) is diverges

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integral from 0 to 4 of sec(x)tan(x)

\int_{0}^{4} sec (x) tan (x) d x

diverges

Solution steps

\int_{0}^{4} sec (x) tan (x) d x

Take the denominator(s) of

sec (x) tan (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}} sec (x) tan (x) d x + \int_{\frac{π}{2}}^{4} sec (x) tan (x) d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

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

\int_{0}^{4} 4 d x

\int_{2}^{3} \frac{1}{x ^{2} + x - 2} d x

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

What is the integral from 0 to 4 of sec(x)tan(x) ?
The integral from 0 to 4 of sec(x)tan(x) is diverges