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
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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)