求定积分-1到1 (x+x^2)/根号下1-x^2 dx 答案是π/2
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解决时间 2021-02-20 03:37
- 提问者网友：孤山下
- 2021-02-19 16:08
怎么做的？
最佳答案
- 五星知识达人网友：逃夭
- 2021-02-19 16:45
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
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
全部回答
- 1楼网友：举杯邀酒敬孤独
- 2021-02-19 18:06
设x = siny，dx = cosy dy
当x = 0，y = 0；当x = 1/2，y = π/6
∫(0→1/2) √(1 - x²) dx
= ∫(0→π/6) √(1 - sin²y) • cosy dy
= ∫(0→π/6) cos²y dy
= (1/2)∫(0→π/6) (1 + cos2y) dy
= (1/2)(y + 1/2 • sin2y) |(0→π/6)
= (1/2)(π/6 + √3/4)
= (3√3 + 2π)/24
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