Question

高等数学解答

Answer 1

问：高等数学解答

答：要求解的积分是 ∫[-π/2, π/2] (sin(x)cos(x)) / (1 + sin^2(x)cos^2(x)) dx。我们可以对被积函数进行因式分解，得到：(sin(x)cos(x))/(1 + sin^2(x)cos^2(x)) = (sin(x)cos(x))/(1 + (sin(x)cos(x))^2)然后，进行变量替换，令 u = sin(x)cos(x)，则 du = (cos^2(x) - sin^2(x)) dx。当 x = -π/2 时，u = 0；当 x = π/2 时，u = 0。因此，积分变为 ∫[0, 0] u/(1 + u^2) du，在积分区间中 u 始终为 0，所以积分结果为 0。总结，∫[-π/2, π/2] (sin(x)cos(x)) / (1 + sin^2(x)cos^2(x)) dx = 0。

答：令u=x^(1/3), 则 x=u^3, dx=3u^2du,于是∫e^[x^(1/3)]dx=3∫u^2*e^udu=3∫u^2de^u=3u^2*e^u-6∫ue^udu=3u^2*e^u-6∫ude^u=3u^2*e^u-6ue^u+6∫e^udu=3(u^2-2u+2)e^u+C=3[x^(2/3)-2x^(1/3)+2]e^[x^(1/3)]+C.