1个回答
2017-11-22
展开全部
换元法令x=3/2sint,t∈[-0.5π,0.5π]
带入后得到
∫(1-x)/[√(9-4x^2)]dx=∫(1-1.5sint)1.5costdt/3cost
=∫(1-1.5sint)0.5dt
=0.5t+0.75cost+C=0.5arcsin2/3x+1/4√9-4x^2+C
带入后得到
∫(1-x)/[√(9-4x^2)]dx=∫(1-1.5sint)1.5costdt/3cost
=∫(1-1.5sint)0.5dt
=0.5t+0.75cost+C=0.5arcsin2/3x+1/4√9-4x^2+C
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