[微积分] 1道

(π/2)
∫ cos (x) / [ (1+sin(x))^(1/2) dx
(0)

(π/2)
=∫ 1 / [ (1+sin(x))^(1/2) dcos（x）
(0)

(π/2)
＝2(1+sin(x))^(1/2)|
(0)
＝2（根号2－1）

(π/2)
∫ cos (x) / [ √(1+sin(x)) dx
(0)

=(π/2)
∫ cos (x) / [ √(1+2sin(x/2)*cos(x/2)) dx
(0)

=(π/2)
∫ cos (x) / [ √(cos^2(x/2)+sin(x/2))^2 dx
(0)

=(π/2)
∫ [cos^2 (x/2)-sin^2(x/2)] / ((cos(x/2)+sin(x/2)) dx
(0)

=(π/2)
∫ [cos (x/2)-sin(x/2)] dx
(0)

明白了吧 哈哈