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trantien.hocmai

$\int_0^\frac{\pi}{2}(sin^4x)dx=\int_0^\frac{\pi}{2}(\frac{1-cos2x}{2})^2dx$
$=\int_0^\frac{\pi}{2}(\frac{cos^22x-2cos2x+1}{4})dx$
$=\int_0^\frac{\pi}{2}(\frac{1+cos4x}{8}-\frac{1}{2}cos2x+\frac{1}{4})dx$
$=(\frac{1}{8}x+\frac{1}{32}sin4x-\frac{1}{4}sin2x+\frac{1}{4}x)|_0^\frac{\pi}{2}$
$=\frac{3\pi}{16}$
 
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