∫¹(1-x²)∧(2/3)dx 0 用换元法,
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/25 12:19:09
∫¹(1-x²)∧(2/3)dx 0 用换元法,
∫¹(1-x²)∧(2/3)dx
0
∫¹(1-x²)∧(2/3)dx
0
∫ [0-->1] (1-x^2)^(3/2) dx
令x=sinu,则dx=cosudu,x:0-->1,u:0-->π/2,此时所有三角函数为正,(1-x^2)^(3/2)=(cosu)^3
代入原式=∫ [0-->π/2] (cosu)^3*cosu du
=∫ [0-->π/2] (cosu)^4 du
=∫ [0-->π/2] (1/2(1+cos2u))^2 du
=1/4∫ [0-->π/2] (1+cos2u)^2 du
=1/4∫ [0-->π/2] (1+2cos2u+(cos2u)^2) du
=1/4∫ [0-->π/2] (1+2cos2u+1/2(1+cos4u)) du
=1/4∫ [0-->π/2] (3/2+2cos2u+1/2cos4u) du
=1/4 (3/2u+sin2u+1/8sin4u) [0-->π/2]
=3π/16
令x=sinu,则dx=cosudu,x:0-->1,u:0-->π/2,此时所有三角函数为正,(1-x^2)^(3/2)=(cosu)^3
代入原式=∫ [0-->π/2] (cosu)^3*cosu du
=∫ [0-->π/2] (cosu)^4 du
=∫ [0-->π/2] (1/2(1+cos2u))^2 du
=1/4∫ [0-->π/2] (1+cos2u)^2 du
=1/4∫ [0-->π/2] (1+2cos2u+(cos2u)^2) du
=1/4∫ [0-->π/2] (1+2cos2u+1/2(1+cos4u)) du
=1/4∫ [0-->π/2] (3/2+2cos2u+1/2cos4u) du
=1/4 (3/2u+sin2u+1/8sin4u) [0-->π/2]
=3π/16
x-9/[(根号)x]+3 dx ∫ x+1/[(根号)x] dx ∫ [(3-x^2)]^2 dx
不定积分 ∫1/(x²-3x+2)dx
∫dx/(1+√(1-2x-x²)) 不定积分 ∫tan-¹(√(a-x)/(a+x))dx 从0积
∫dx/(2x-3)²
计算 ∫(x^4-2x^3+x^2+1)/x(x-1)² dx
∫dx/(1-x∧2)∧3/2
广义积分求解∫ 1/x²-4x+3 dx(0到2)∫1/x(lnx)² dx (0到无穷)
∫1/(2x+3)²dx的不定积分
∫(2^x+3^x)²dx
∫2 -1|x²-x|dx
∫(2x)/(1+x²)dx
∫x²/(x-1)dx