Can someone please help me find the integral of 6x^2(cos^3x)
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just do an integration by parts { u = 6 x² , dv = (1 - sin²x ) cos x dx }
to get to [ 6x² sin x - 2 x² sin^3 x+ 4x cos^3 x - 4 sin x+ ( 4 / 3 ) sin^3 x ]
check out my work .
to get to [ 6x² sin x - 2 x² sin^3 x+ 4x cos^3 x - 4 sin x+ ( 4 / 3 ) sin^3 x ]
check out my work .