Question

求一个不定积分的题目，谢谢

Answer 1

采用换元法，设x=sint ，则原式变为

积分(Sintcost)^2 dsint=积分(sint)^2 (cost)^3 dt

然后采用分部积分就可求出了

Answer 2

设 x = sint，则 dx = cost*dt
∫x^2*√(1-x^2) *dx
=∫(sint)^2*cost *(cost)*dt
=∫(sint*cost)^2*dt
=1/4*∫(2sint*cost)^2 *dt
=1/4*∫(sin2t)^2 *dt
=1/8*∫2*(sin2t)^2 *dt
=1/8*∫[1-cos4t]*dt
=1/8*∫dt - 1/8*∫cos4t*dt
=1/8*t -1/32*sin4t + C
=1/8*t - 1/16* sin2t*cos2t + C
=1/8*t - 1/8 * sint*cost*[1-2(sint)^2] + C
=1/8*t - 1/8*x*√(1-x^2)*(1-2x^2) + C