The value of e ^ - (π)/(4) + ∫ _ 0 ^ (π)/(4) e ^ - x tan ^ 50 xdx ∫...

Let,

$I = \dfrac{{e}^{- \dfrac{\pi}{4}} + {∫}_{0}^{\dfrac{\pi}{4}}{e}^{- x}tan{}^{50}xdx}{{∫}_{0}^{\dfrac{\pi}{4}}{e}^{- x}\left(\tan^{49}x + \tan^{51}x\right)dx}$

Now let ${I}_{1} = {∫}_{0}^{\dfrac{\pi}{4}}{e}^{- x} ( \tan^{49}x + \tan^{51}x ) dx$

Now solving, ${I}_{2} = {∫}_{0}^{\dfrac{\pi}{4}}{e}^{- x}\tan^{50}xdx$ using byparts we get,

${I}_{2} = {∫}_{0}^{\dfrac{\pi}{4}}{e}^{- x}\tan^{50}xdx$

$⇒ {I}_{2} = {\left[- {e}^{- x}\tan^{50}x\right]}_{0}^{\dfrac{\pi}{4}} - {∫}_{0}^{\dfrac{\pi}{4}}50\tan^{49}x{sec}^{2}x\left(- {e}^{- x}\right)dx$

$⇒ {I}_{2} = - {e}^{- \dfrac{\pi}{4}} + {∫}_{0}^{\dfrac{\pi}{4}}50\tan^{49}x\left(1 + \tan^{2}x\right){e}^{- x}dx$

$⇒ {I}_{2} = - {e}^{- \dfrac{\pi}{4}} + 50{∫}_{0}^{\dfrac{\pi}{4}}\left(\tan^{49}x + \tan^{51}x\right){e}^{- x}dx$

$⇒ {I}_{2} = - {e}^{- \dfrac{\pi}{4}} + 50{I}_{1}$

$⇒ {∫}_{0}^{\dfrac{\pi}{4}}{e}^{- x}\tan^{50}xdx + {e}^{- \dfrac{\pi}{4}} = 50{I}_{1}$

Now putting the given integral $I$ we get,

$I = \dfrac{{e}^{- \dfrac{\pi}{4}} + {∫}_{0}^{\dfrac{\pi}{4}}{e}^{- x}tan{}^{50}xdx}{{∫}_{0}^{\dfrac{\pi}{4}}{e}^{- x}\left(\tan^{49}x + \tan^{51}x\right)dx}$

$⇒ I = \dfrac{50{I}_{1}}{{I}_{1}} = 50$