What is the integral from 0 to pi/6 of 1/(cos^2(4x)) ?

The integral from 0 to pi/6 of 1/(cos^2(4x)) is diverges

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integral from 0 to pi/6 of 1/(cos^2(4x))

\int_{0}^{\frac{π}{6}} \frac{1}{cos ^{2} ( 4 x )} d x

diverges

Solution steps

\int_{0}^{\frac{π}{6}} \frac{1}{cos ^{2} ( 4 x )} d x

Take the denominator(s) of

\frac{1}{cos ^{2} ( 4 x )}

and compare to zero:

x = \frac{π}{8}

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}{8}} \frac{1}{cos ^{2} ( 4 x )} d x + \int_{\frac{π}{8}}^{\frac{π}{6}} \frac{1}{cos ^{2} ( 4 x )} d x

\int_{0}^{\frac{π}{8}} \frac{1}{cos ^{2} ( 4 x )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to pi/6 of 1/(cos^2(4x)) ?
The integral from 0 to pi/6 of 1/(cos^2(4x)) is diverges