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Ta có
$e^{\sin ^{2}x} \sin 4x$
$=e^{\sin ^{2}x} (2\sin 2x\cos 2x)$
$=-2e^{\sin ^{2}x} (2\sin 2x) + (\cos 2x +2)e^{\sin ^{2}x} (2\sin 2x)$
$=e^{\sin ^{2}x} (\cos 2x +2)' + (\cos 2x +2)(e^{\sin ^{2}x} )'$
$=(e^{\sin ^{2}x} (\cos 2x +2))' $
Vậy
$ \int\limits_{0}^{\frac{\pi}{2}} e^{\sin ^{2}x} \sin 4xdx =\left[ {e^{\sin ^{2}x} (\cos 2x +2)} \right]_{0}^{\frac{\pi}{2}}=\boxed{2(e-3)}$
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