∫1(上标)-1(下标)√(1-x²)dx=?
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∫1(上标)-1(下标)√(1-x²)dx=?
∫1(上标)-1(下标)√(1-x²)dx=?
∫1(上标)-1(下标)√(1-x²)dx=?
设x=sint,因此(-π/2≤t≤π/2)
则原式变为∫_(-π/2)^(π/2)▒cost d_sint =∫_(-π/2)^(π/2)▒〖(cost )^2 〗 d_t
=∫_(π/(-2))^(π/2)▒(1+cos2t)/2 d_t
=(t/2+sin2t/4)|_((-π)/2)^(π/2)
=π/2