Mathematics

# $I = \dfrac {2}{\pi}\int_{-\pi/4}^{\pi/4} \dfrac {dx}{(1 + e^{\sin x})(2 - \cos 2x)}$ then find $27I^{2}$ equals ________.

4.00

##### SOLUTION
$I = \dfrac {2}{\pi}\int_{-\dfrac {\pi}{4}^{\pi/4}} \dfrac {dx}{(1 + e^{\sin x})(2 - \cos 2x)} ......(1)$
by $a + b - x$ property
$I = \dfrac {2}{\pi} \int_{-\dfrac {\pi}{4}}^{\pi/4} \dfrac {dx}{(1 + e^{-sin x})(2 - cos 2x)} = \dfrac {2}{\pi} \int_{-\dfrac {\pi}{4}}^{\pi/4} \dfrac {e^{\sin xdx}}{(1 + e^{\sin x})(2 - \cos 2x)}dx .... (2)$
adding (1) and (2)
$2I = \dfrac {2}{\pi} \int_{-\dfrac {\pi}{4}}^{\pi/4} \dfrac {dx}{(1 + e^{-\sin x})(2 - \cos 2x)} = \dfrac {2}{\pi} \int_{-\dfrac {\pi}{4}}^{\pi/4} \dfrac {e^{\sin x}dx}{(1 + e^{\sin x})(2 - \cos 2x)}dx ....(2)$
put $\tan x = t, \sec^{2} x dx = dt$
$= \dfrac {2}{\pi}\int_{0}^{1} \dfrac {dt}{3t^{2} + 1} = \dfrac {2}{3\pi} \dfrac {1}{\left (\dfrac {1}{\sqrt {3}}\right )} \left (\tan^{-1} \left (\dfrac {t}{1/\sqrt {3}}\right )\right )^{1}_{0} = \dfrac {2}{\sqrt {3}\pi} (\tan^{-1}(\sqrt {3}) - \tan^{-1} (0)) = \dfrac {2}{\sqrt {3}\pi} \left (\dfrac {\pi}{3}\right ) = \dfrac {2}{3\sqrt {3}}$
Now $27I^{2} = 27\times \dfrac {4}{27} = 4$.

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