What is the integral from 1 to 2 of sec^3(θ) ?

The integral from 1 to 2 of sec^3(θ) is diverges

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integral from 1 to 2 of sec^3(θ)

\int_{1}^{2} sec^{3} (θ) d θ

diverges

Solution steps

\int_{1}^{2} sec^{3} (θ) d θ

Take the denominator(s) of

sec^{3} (θ)

and compare to zero:

θ = \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_{1}^{\frac{π}{2}} sec^{3} (θ) d θ + \int_{\frac{π}{2}}^{2} sec^{3} (θ) d θ

\int_{1}^{\frac{π}{2}} sec^{3} (θ) d θ =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

What is the integral from 1 to 2 of sec^3(θ) ?
The integral from 1 to 2 of sec^3(θ) is diverges