Answer to Question #247656 in Calculus for JaytheCreator

Question #247656

Show that "\\int" π‘₯ 3 cos π‘₯ 𝑑π‘₯ πœ‹ 2 0 = πœ‹ 3 8 βˆ’ 3πœ‹ + 6


1
Expert's answer
2021-10-06T18:12:45-0400

"\\int_0^{2 \\pi} x^3\\cos x dx= x^3\\sin x\\Biggr|_0^{2\\pi}-\\int_0^{2 \\pi}3x^2 \\sin x dx\\\\\n=-3\\left[-x^2 \\cos x\\Biggr|_0^{2 \\pi}+2\\int_0^{2 \\pi}x \\cos x dx\\right]\\\\\n=-3\\left[-4 \\pi^2+2\\left(x \\sin x\\Biggr|_0^{2\\pi}-\\int_0^{2\\pi}\\sin x dx\\right)\\right]\\\\\n=-3[-4\\pi^2+2(\\cos x\\Biggr|_0^{2\\pi})]\\\\\n=-3[-4\\pi^2+2]=12\\pi^2-6"


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