What is the integral from pi/2 to pi of sec^2(8x) ?

The integral from pi/2 to pi of sec^2(8x) is diverges

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integral from pi/2 to pi of sec^2(8x)

\int_{\frac{π}{2}}^{π} sec^{2} (8 x) d x

diverges

Solution steps

\int_{\frac{π}{2}}^{π} sec^{2} (8 x) d x

Take the denominator(s) of

sec^{2} (8 x)

and compare to zero:

x = \frac{9 π}{16}, x = \frac{11 π}{16}, x = \frac{13 π}{16}, x = \frac{15 π}{16}

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_{\frac{π}{2}}^{\frac{9 π}{16}} sec^{2} (8 x) d x + \int_{\frac{9 π}{16}}^{\frac{11 π}{16}} sec^{2} (8 x) d x + \int_{\frac{11 π}{16}}^{\frac{13 π}{16}} sec^{2} (8 x) d x + \int_{\frac{13 π}{16}}^{\frac{15 π}{16}} sec^{2} (8 x) d x + \int_{\frac{15 π}{16}}^{π} sec^{2} (8 x) d x

\int_{\frac{π}{2}}^{\frac{9 π}{16}} sec^{2} (8 x) d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 4}^{1} (x - 4) d x

\int_{0}^{2 π} 1 + sin (x) d x

\int_{0}^{\frac{π}{2}} 2 cos (θ) d θ

\int_{0}^{4} x^{2} + 4 d x

\int_{- 8}^{0} 3 x d x

What is the integral from pi/2 to pi of sec^2(8x) ?
The integral from pi/2 to pi of sec^2(8x) is diverges