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

∫02Ο€x3cos⁑xdx=x3sin⁑x∣02Ο€βˆ’βˆ«02Ο€3x2sin⁑xdx=βˆ’3[βˆ’x2cos⁑x∣02Ο€+2∫02Ο€xcos⁑xdx]=βˆ’3[βˆ’4Ο€2+2(xsin⁑x∣02Ο€βˆ’βˆ«02Ο€sin⁑xdx)]=βˆ’3[βˆ’4Ο€2+2(cos⁑x∣02Ο€)]=βˆ’3[βˆ’4Ο€2+2]=12Ο€2βˆ’6\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\\ =-3\left[-x^2 \cos x\Biggr|_0^{2 \pi}+2\int_0^{2 \pi}x \cos x dx\right]\\ =-3\left[-4 \pi^2+2\left(x \sin x\Biggr|_0^{2\pi}-\int_0^{2\pi}\sin x dx\right)\right]\\ =-3[-4\pi^2+2(\cos x\Biggr|_0^{2\pi})]\\ =-3[-4\pi^2+2]=12\pi^2-6


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