d〔∫下限a,上限b cos^2 dx〕＝?怎么算,

网友回答

∫(a~b) cos²x dx
= (1/2)∫ (1 + cos2x) dx
= (1/2)(x + 1/2 * sin2x) |(a~b)
= (1/2)(b + 1/2 * sin2b) - (1/2)(a + 1/2 * sin2a)
= (1/2)b + (1/2)sinbcosb - (1/2)a - (1/2)sinacosa
= (1/2)(b - a) + (1/2)(sinbcosb - sinacosa)