Question

`\red{ \rm\int\limits_(0)^{ (\pi)/(2)} l {n}^(2) \bigg( \frac{ {e}^{ - {x}^(2) } }{ \cos(x) } (1 + \cos(4x)) \bigg ) dx } \\`​

Answer 1

First we rewrite

`(1 + \cos(4x))/(\cos(x)) = 2\cos^2(2x)`

then expand the integrand as

`\displaystyle \ln^2(2) - 2 \ln(2) x^2 + x^4 \\\\ {} + 2\ln(2) \ln(\cos^2(2x)) - 2\ln(2) \ln(\cos(x)) - 2x^2 \ln(\cos^2(2x) + 2x^2 \ln(\cos(2x)) \\\\ {} + \ln^2(\cos^2(2x)) + \ln^2(\cos(x)) - 2 \ln(\cos(x)) \ln(\cos^2(2x))`

We'll use the following identity:

`\displaystyle \cos(2kx) = \frac{e^(i2kx) + e^(-i2kx)}2 \\\\ \sum_(k=1)^\infty \frac{\cos(2kx)}k = \frac12 \left(\sum_(k=1)^\infty \frac{(e^(i2x))^k}k + \frac{(e^(-i2x))^k}k\right) \\\\ \sum_(k=1)^\infty \frac{\cos(2kx)}k = -\frac12 \left(\ln(1-e^(i2kx)) + \ln(1 - e^(-i2kx))\right) \\\\ \implies \ln(\sin(x)) = -\ln(2) - \sum_(k=1)^\infty \frac{\cos(2kx)}k`

as well as the fact that for any integer n,

`\displaystyle \int_0^(\frac\pi2) \cos(2nx) \, dx = 0`

Consult the attachments for the integrals of the non-trivial terms.

Putting everything together, the end result is then

`\displaystyle \int_0^(\frac\pi2) \ln^2\left((e^(-x^2))/(\cos(x))(1+\cos(4x))\right) \, dx \\\\ = \boxed{(\pi^5)/(160) + \frac{\pi^3}4 - (11\pi)/(16) \zeta(3)}`