大一微积分求解①∫cos²xsin²xdx②∫cot²x+cos²(x/2)dx
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![大一微积分求解①∫cos²xsin²xdx②∫cot²x+cos²(x/2)dx](/uploads/image/z/6092051-59-1.jpg?t=%E5%A4%A7%E4%B8%80%E5%BE%AE%E7%A7%AF%E5%88%86%E6%B1%82%E8%A7%A3%E2%91%A0%E2%88%ABcos%26%23178%3Bxsin%26%23178%3Bxdx%E2%91%A1%E2%88%ABcot%26%23178%3Bx%EF%BC%8Bcos%26%23178%3B%EF%BC%88x%EF%BC%8F2%29dx)
大一微积分求解①∫cos²xsin²xdx②∫cot²x+cos²(x/2)dx
大一微积分求解①∫cos²xsin²xdx②∫cot²x+cos²(x/2)dx
大一微积分求解①∫cos²xsin²xdx②∫cot²x+cos²(x/2)dx
∫ cos²xsin²x dx
= ∫ [(1/2)sin2x]² dx
= (1/4)∫ sin²2x dx
= (1/8)∫ [1 - cos4x] dx
= x/8 - (1/32)sin4x + C
∫ [cot²x + cos²(x/2)] dx
= ∫ [csc²x - 1 + (1 + cosx)/2] dx
= - cotx - x + x/2 + (1/2)sinx + C
= - cotx - x/2 + (1/2)sinx + C
都很简单,嘿嘿