# Iron Front Liberation 1944 Crack 1.05 WORK Iron Front Liberation 1944 Crack 1.05

i got the new patch for iron front 1.05 and tried it my mic wont work and when i restart it and launch. those made possible during the d-day bombardment. english languageQ: Application of the mean value theorem for the general integral Consider the integral $$f(x) = \int^{\frac{2\pi}{3}}_{0}\cos^{n-1}(x)-\sin^{n-1}(x) \cdot \cos(ax) \cdot {1 \over \sqrt{1+a^2}} \; d\theta$$ The issue is as to how to evaluate the integral by applying the mean value theorem; I have tried every method. My work so far: $$f(x) =\sum^n_{r=0} a_r\int^{\frac{2\pi}{3}}_{0}\cos^{n-1-r}(x)-\sin^{n-1-r}(x) \cdot \cos(ax) \cdot {1 \over \sqrt{1+a^2}} \; d\theta$$ where $a_r$ is $$a_r = {(-1)^{n-r} \over \sqrt{1+a^2}}$$ With the mean value theorem we can conclude that $f'(c)$ exists for some $c \in [0,1/2\pi]$. This leads me to evaluating the integrals $$\int^{\frac{2\pi}{3}}_{0}\cos^{n-1-r}(x)-\sin^{n-1-r}(x) \cdot \cos(ax) \cdot {1 \over \sqrt{1+a^2}} \; d\theta$$ Which, according to WolframAlpha, is simply $$\int^{\frac{2\pi}{3}}_{0}e^{inx} – e^{ -inx} {1 \over \sqrt{1+a^2}} \; d\theta$$ We can try to express the denominator as a ratio of complex exponentials, but the denominator, when I replace $a$ with $i$, is a product of an even number of complex exponential terms, which poses a problem for me