The value of $${{{e^{ - {\pi \over 4}}} + \int\limits_0^{{\pi \over 4}} {{e^{ - x}}{{\tan }^{50}}xdx} } \over {\int\limits_0^{{\pi \over 4}} {{e^{ - x}}({{\tan }^{49}}x + {{\tan }^{51}}x)dx} }}$$ is

<p>We&#39;re given the expression:</p> <p>$$\frac{e^{-\pi/4} + \int_0^{\pi / 4} e^{-x} \tan ^{50} x dx}{\int_0^{\pi / 4} e^{-x}(\tan x)^{49} dx + \int_0^{\pi / 4} e^{-x}(\tan x)^{51} dx}$$</p> <p>Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x} \tan^n x$ from 0 to $\pi / 4$, where $n$ is an integer. Let&#39;s denote this integral as $I(n)$:</p> <p>$I(n) = \int_0^{\pi / 4} e^{-x} \tan^n x dx$</p> <p>We can then rewrite the original expression in terms of $I(n)$:</p> <p>$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)}$</p> <p>Now, we&#39;ll apply the method of integration by parts, which states that for two functions $u(x)$ and $v(x)$:</p> <p>$\int u dv = uv - \int v du$</p> <p>We&#39;ll choose:</p> <p>$u = \tan^n x, \quad dv = e^{-x} dx$</p> <p>Then we get:</p> <p>$du = n \tan^{n-1} x \sec^2 x dx, \quad v = -e^{-x}$</p> <p>Applying integration by parts, we have:</p> <p>$$I(n) = -e^{-x} \tan^n x \Bigg|_0^{\pi / 4} + n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x \sec^2 x dx$$</p> <p>Since $\tan(\pi / 4) = 1$, the first term evaluates to:</p> <p>$-e^{-\pi / 4}$</p> <p>The second term becomes:</p> <p>$$n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x (1 + \tan^2 x) dx = n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x dx + n \int_0^{\pi / 4} e^{-x} \tan^{n+1} x dx$$</p> <p>This is equal to:</p> <p>$n(I(n-1) + I(n+1))$</p> <p>So we have:</p> <p>$I(n) = -e^{-\pi / 4} + n(I(n-1) + I(n+1))$</p> <p>Now we can substitute $n = 50$ into this equation:</p> <p>$I(50) = -e^{-\pi / 4} + 50(I(49) + I(51))$</p> <p>So the original expression becomes:</p> <p>$$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)} = \frac{e^{-\pi/4} - e^{-\pi / 4} + 50(I(49) + I(51))}{I(49) + I(51)} = 50$$</p>