求不定积分∫(1-x)dx/√(9-4x²) 答案是(1/2)arcsin(2x/3)+√(9-4x²)/4+C我算出来的是(1/2)arcsin(2x/3)+√(9-4x²)/8+C就差一个数字帮忙看看哪个错了?
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![求不定积分∫(1-x)dx/√(9-4x²) 答案是(1/2)arcsin(2x/3)+√(9-4x²)/4+C我算出来的是(1/2)arcsin(2x/3)+√(9-4x²)/8+C就差一个数字帮忙看看哪个错了?](/uploads/image/z/2541892-4-2.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%281-x%29dx%2F%E2%88%9A%289-4x%26sup2%3B%29+%E7%AD%94%E6%A1%88%E6%98%AF%281%2F2%29arcsin%282x%2F3%29%2B%E2%88%9A%289-4x%26sup2%3B%29%2F4%2BC%E6%88%91%E7%AE%97%E5%87%BA%E6%9D%A5%E7%9A%84%E6%98%AF%281%2F2%29arcsin%282x%2F3%29%2B%E2%88%9A%289-4x%26sup2%3B%29%2F8%2BC%E5%B0%B1%E5%B7%AE%E4%B8%80%E4%B8%AA%E6%95%B0%E5%AD%97%E5%B8%AE%E5%BF%99%E7%9C%8B%E7%9C%8B%E5%93%AA%E4%B8%AA%E9%94%99%E4%BA%86%3F)
求不定积分∫(1-x)dx/√(9-4x²) 答案是(1/2)arcsin(2x/3)+√(9-4x²)/4+C我算出来的是(1/2)arcsin(2x/3)+√(9-4x²)/8+C就差一个数字帮忙看看哪个错了?
求不定积分∫(1-x)dx/√(9-4x²)
答案是(1/2)arcsin(2x/3)+√(9-4x²)/4+C
我算出来的是(1/2)arcsin(2x/3)+√(9-4x²)/8+C
就差一个数字
帮忙看看哪个错了?
求不定积分∫(1-x)dx/√(9-4x²) 答案是(1/2)arcsin(2x/3)+√(9-4x²)/4+C我算出来的是(1/2)arcsin(2x/3)+√(9-4x²)/8+C就差一个数字帮忙看看哪个错了?
∫(1-x)dx/√(9-4x²)
=∫1/√(9-4x²)dx-∫x/√(9-4x²)dx
=[∫1/√(9/4-x²)dx]/2-[∫1/√(9-4x²)dx²]/2
=∫1/√(9-4x²)dx+[∫1/√(9-4x²)d(9-4x²)]/8
=arcsin(2x/3)/2+√(9-4x²)/4+C
∫1/√(9-4x²)d(9-4x²)
=∫(9-4x²)^(-1/2)d(9-4x²)
=[1/(1-1/2)](9-4x²)^(1-1/2)
=2√(9-4x²)
你这里漏掉了2
∫(1-x)dx/√(9-4x²)
设 2x/3 = sint, 即 t = arcsin(2x/3)
∫(1-(3sint)/2)d(3sint/2)/√(9-9sin²t)
= ∫(1-(3sint)/2)d(3sint/2)/3cost
= (1/2) ∫(1-(3sint)/2)/cost * d(sint)
= (1/2) ∫(...
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∫(1-x)dx/√(9-4x²)
设 2x/3 = sint, 即 t = arcsin(2x/3)
∫(1-(3sint)/2)d(3sint/2)/√(9-9sin²t)
= ∫(1-(3sint)/2)d(3sint/2)/3cost
= (1/2) ∫(1-(3sint)/2)/cost * d(sint)
= (1/2) ∫(1-(3sint)/2) dt
= (1/2) (∫dt-∫(3sint)/2) dt)
= (1/2) (∫dt- (3/2)∫(sint)dt)
= (1/2) (t + (3/2)cost) + C
= (1/2) t + (3/4)√(1-sin²t) + C
= (1/2)arcsin(2x/3) + √(9-4x²)/4 + C
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