Question

求定积分-1到1 (x+x^2)/根号下1-x^2 dx 答案是π/2

怎么做的？

Answer 1

解：
令x=sint，则t=arcsinx
x：-1→1，则t：-π/2→π/2
∫[-1：1][(x+x²)/√(1-x²)]dx
=∫[-1：1][x/√(1-x²)]dx +∫[-1：1][x²/√(1-x²)]dx
=0+2∫[0：π/2][sin²t/√(1-sin²t)]d(sint)
=2∫[0：π/2](sin²t·cost/cost)dt
=∫[0：π/2](1-cos2t)dt
=(t-½sin2t)|[0：π/2]
=(π/2 -½sinπ)-(0-½sin0)
=π/2-0 -0+0
=π/2