求∫xcos^2xdx等于多少,上限是2π,下限是0.

Question

求∫xcos^2xdx等于多少,上限是2π,下限是0.

Answer 1

∫(0→2π) xcos²x dx
= ∫(0→2π) x • (1 + cos2x)/2 dx
= (1/2)∫(0→2π) x dx + (1/2)∫(0→2π) xcos2x dx
= (1/4)[ x² ]：(0→2π) + (1/4)∫(0→2π) x d(sin2x)
= (1/4)(4π²) + (1/4)xsin2x：(0→2π) - (1/4)∫(0→2π) sin2x dx
= π² + 0 + (1/8)cos2x：(0→2π)
= π² + (1/8)(1 - 1)
= π²