Question

don't post irrelevant answers

Answer 1

✪ᴀɴsᴡᴇʀ✪

$\star \star \blue{\boxed{\red{\boxed{27I² = 4}}}} \star \star$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

᯽ғᴏʀᴍᴜʟᴀ᯽

$\int_{a}^{b}f(x)dx = \int_{a}^{b}f((a+b) - x)dx$

☆ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ☆

$I = \frac{2}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{dx}{(1+{e}^{sin(x)})(2-cos(2x) )}.... (1)$

Using the above formula

$I = \frac{2}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{dx}{(1+{e}^{sin[(-\frac{\pi}{4} + \frac{\pi}{4}) - x]})(2-cos[2[(-\frac{\pi}{4} +\frac{\pi}{4}) - x]] )}$

$I = \frac{2}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{dx}{(1+{e}^{sin[0 - x]})(2-cos[2[0 - x]] )}$

$I = \frac{2}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{dx}{(1+{e}^{-sin(x)})(2-cos(2x) )}$

$I = \frac{2}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{{e}^{sin(x)}dx}{{e}^{sin(x)}(1+{e}^{-sin(x)})(2-cos(2x) )}$

$I = \frac{2}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{{e}^{sin(x)}dx}{({e}^{sin(x)} + 1)(2-cos(2x) )}.... (2)$

Adding (1) and (2), we get

$2I = \frac{2}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{dx}{(1+{e}^{sin(x)})(2-cos(2x)))} + \frac{2}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{{e}^{sin(x)}dx}{({e}^{sin(x)} + 1)(2-cos(2x) )}$

$2I = \frac{2}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{(1+{e}^{sin(x)})dx}{(1+{e}^{sin(x)})(2-cos(2x))}$

$I = \frac{1}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{dx}{2-cos(2x)}$

$I = \frac{1}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{dx}{2 - \frac{1-{tan}^{2}(x)} {1+{tan}^{2}(x)} }$

$I = \frac{1}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{[1+{tan}^{2}(x)]dx}{2[1+{tan}^{2}(x)] -[1-{tan}^{2}(x)] }$

$I = \frac{1}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{[{sec}^{2}(x)]dx}{2+2{tan}^{2}(x) - 1 + {tan}^{2}(x)}$

$I = \frac{1}{π}\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}} \frac{[{sec}^{2}(x)]dx}{3{tan}^{2}(x) + 1}$

Substitute $tan(x) = t$

$\implies {sec}^{2}(x)dx = dt$

As $x \to -\frac{\pi}{4}, t \to - 1$

As $x \to \frac{\pi}{4}, t \to 1$

$I = \frac{1}{π}\int_{-1 }^{1} \frac{dt}{3{t}^{2} + 1}$

$I = \frac{1}{π}\int_{-1 }^{1} \frac{dt}{{(\sqrt{3}t) }^{2}+ 1}$

$I = \frac{1}{π}[\frac{1}{\sqrt{3}}{tan}^{-1}(\sqrt{3}t) ]_{-1 }^{1}$

$I = \frac{1}{\sqrt{3}\pi }[{tan}^{-1}(\sqrt{3}) -{tan}^{-1}(-\sqrt{3}) ]$

$I = \frac{1}{\sqrt{3}\pi }[{tan}^{-1}(\sqrt{3}) +{tan}^{-1}(\sqrt{3}) ]$

$I = \frac{1}{\sqrt{3}\pi }[\frac{\pi }{3} +\frac{\pi } {3}]$

$I = \frac{1}{\sqrt{3}\pi }\frac{2\pi }{3}$

$I = \frac{2}{3\sqrt{3 }}$

${I}^{2} = \frac{4}{9\times 3}$

$27{I}^{2} = 4$

Answer 2

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