作业帮定积分的两条题目?计算啊区间是-0.5到0.5 【（x*x）/（根号（1-x*x））】dx 区间是0到2 [1/（根号（1+2*x*x））]dx以上2个题用定积分的换元法做
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 13:25:11
![定积分的两条题目?计算啊区间是-0.5到0.5 【（x*x）/（根号（1-x*x））】dx 区间是0到2 [1/（根号（1+2*x*x））]dx以上2个题用定积分的换元法做](/uploads/image/z/938847-39-7.jpg?t=%E5%AE%9A%E7%A7%AF%E5%88%86%E7%9A%84%E4%B8%A4%E6%9D%A1%E9%A2%98%E7%9B%AE%3F%E8%AE%A1%E7%AE%97%E5%95%8A%E5%8C%BA%E9%97%B4%E6%98%AF-0.5%E5%88%B00.5+%E3%80%90%EF%BC%88x%2Ax%EF%BC%89%2F%EF%BC%88%E6%A0%B9%E5%8F%B7%EF%BC%881-x%2Ax%EF%BC%89%EF%BC%89%E3%80%91dx+%E5%8C%BA%E9%97%B4%E6%98%AF0%E5%88%B02+%5B1%2F%EF%BC%88%E6%A0%B9%E5%8F%B7%EF%BC%881%2B2%2Ax%2Ax%EF%BC%89%EF%BC%89%5Ddx%E4%BB%A5%E4%B8%8A2%E4%B8%AA%E9%A2%98%E7%94%A8%E5%AE%9A%E7%A7%AF%E5%88%86%E7%9A%84%E6%8D%A2%E5%85%83%E6%B3%95%E5%81%9A)
定积分的两条题目?计算啊区间是-0.5到0.5 【（x*x）/（根号（1-x*x））】dx 区间是0到2 [1/（根号（1+2*x*x））]dx以上2个题用定积分的换元法做
定积分的两条题目?计算啊
区间是-0.5到0.5 【（x*x）/（根号（1-x*x））】dx
区间是0到2 [1/（根号（1+2*x*x））]dx
以上2个题用定积分的换元法做
定积分的两条题目?计算啊区间是-0.5到0.5 【（x*x）/（根号（1-x*x））】dx 区间是0到2 [1/（根号（1+2*x*x））]dx以上2个题用定积分的换元法做
∫[x^2/√(1-x^2)]dx,[-0.5,0.5]
令x=sint,积分范围为[-π/6,π/6]
√(1-x^2)=cost,dx=costdt
∫[x^2/√(1-x^2)]dx,[-0.5,0.5]
=∫(sint)^2dt,[-π/6,π/6]
=∫[1-cos2t]/2dt,[-π/6,π/6]
=t/2-sin2t/4,[-π/6,π/6]
=π/12-√3/8-(-π/12+√3/8)
=π/6-√3/4
∫dx/√(1+2x^2),[0,2]
令√2*x=tant,则积分范围为[0,arctan(2√2)]
√(1+2x^2)=sect,dx=(1/√2)(sect)^2dt
∫dx/√(1+2x^2),[0,2]
=(1/√2)∫sectdt,[0,arctan(2√2)]
=(1/√2)∫dt/cost,[0,arctan(2√2)]
=(1/√2)∫costdt/[1-(sint)^2],[0,arctan(2√2)]
=(1/√2)∫dsint/[1-(sint)^2],[0,arctan(2√2)]
=(1/2√2)∫[1/(1-sint)+1/(1+sint)]dsint,[0,arctan(2√2)]
=(1/2√2)∫[dln[(1+sint)/(1-sint)],[0,arctan(2√2)]
=(1/2√2)ln[(1+sint)/(1-sint)],[0,arctan(2√2)]
=(1/2√2)ln[(1+sinarctan2√2)/(1-sinarctan2√2)]
=(1/2√2)ln[(1+2√2/3)/(1-2√2/3)]
=(1/√2)ln(3+2√2)