推 suhorng :可以用三角代換變成∫sin^2(t)dt 03/23 22:36
→ suhorng :或是分成 -x * -x/√(1-x^2) 分部積分 (差不多) 03/23 22:37
→ suker :令x=siny ; y=arcsinx 03/23 22:39
→ suker :∫sin^2(y)dt =y/2 -sin2y /4 +C 03/23 22:40
→ suker :∫sin^2(y)dy =y/2 -sin2y /4 +C =y/2-sinycosy/2+C 03/23 22:41
→ suker :=arcsinx /2 -{x*√(1-x^2)}/2}+C ,cosy=√(1-x^2) 03/23 22:43
→ kniver999 :要怎麼用三角代換變成那樣？ 03/23 23:00
→ suhorng :完整算式三樓板友寫出來了 03/23 23:06
→ suhorng :y=arcsin(x), so dy = dx/√(1-x^2) 03/23 23:06
→ suhorng :∫x^2 dx/√(1-x^2) = ∫sin^2(y) dy 03/23 23:07
→ kniver999 :我會了 謝謝 03/23 23:53