∫(2-sec^2(x))^1/2 d(sec x) [上限为π/4下限为0]
请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec(t)]^2d所确定,求d^2y/dx^2
∫ sec^2 x dx
∫secx/sec^2x-1 dx
求定积分∫(上限为π/2.下限为0)|1/2-sin x| dx
定积分∫(sinx+sin^3(x))^1/2 [上限为π下限为0]
∫sec^4x dx ∫sec^2x tan^2x dx
若定积分∫f(x)[(x^3 sin^2( x))d(x)/x^4+2x^2+1] [上限为5下限为-5],
定积分 上限为1 下限为0 ∫ (x^2)/(1+x^2)^3 dx
求(sec x)^2积分,
lim(x趋近零)[∫(1+t^2) e^(t^2-x^2)d(x)]/x^2 {定积分上限是x^2,下限为0}
①∫(上限为正无穷,下限为0)1/(x^2+4x+5)dx
∫(上限1,下限0)dx∫(上限1,下限x)x^2*siny^2dy