∫dx/(1+√(1-2x-x²)) 不定积分 ∫tan-¹(√(a-x)/(a+x))dx 从0积到a的定积分值
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∫dx/(1+√(1-2x-x²)) 不定积分 ∫tan-¹(√(a-x)/(a+x))dx 从0积到a的定积分值
∫dx/(1+√(1-2x-x²)) 不定积分 ∫tan-¹(√(a-x)/(a+x))dx 从0积到a的定积分值
∫dx/(1+√(1-2x-x²)) 不定积分 ∫tan-¹(√(a-x)/(a+x))dx 从0积到a的定积分值
其实第一题挺有难度的...
∫ arctan√[(a - x)/(a + x)] dx
= x * arctan√[(a - x)/(a + x)] - ∫ x * (- 1/2) * 1/√(a² - x²) dx
= x * arctan√[(a - x)/(a + x)] + (1/2)∫ x/√(a² - x²) dx
= x * arctan√[(a - x)/(a + x)] + (1/2)(- 1/2)∫ 1/√(a² - x²) d(a² - x²)
= x * arctan√[(a - x)/(a + x)] - (1/4) * 2√(a² - x²) + C
= x * arctan√[(a - x)/(a + x)] - (1/2)√(a² - x²) + C
所以∫(0→a) arctan√[(a - x)/(a + x)] dx
= 0 - [0 - (1/2)√(a²)]
= |a|/2