Show that ∫\int∫ 𝑥 3 cos 𝑥 𝑑𝑥 𝜋 2 0 = 𝜋 3 8 − 3𝜋 + 6
∫02πx3cosxdx=x3sinx∣02π−∫02π3x2sinxdx=−3[−x2cosx∣02π+2∫02πxcosxdx]=−3[−4π2+2(xsinx∣02π−∫02πsinxdx)]=−3[−4π2+2(cosx∣02π)]=−3[−4π2+2]=12π2−6\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∫02πx3cosxdx=x3sinx∣∣02π−∫02π3x2sinxdx=−3⎣⎡−x2cosx∣∣02π+2∫02πxcosxdx⎦⎤=−3⎣⎡−4π2+2⎝⎛xsinx∣∣02π−∫02πsinxdx⎠⎞⎦⎤=−3[−4π2+2(cosx∣∣02π)]=−3[−4π2+2]=12π2−6
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