A Maths一問...

d/dx (sin x)^2 = 2(sinx)(cosx)
Let u=sin x, y=u^2
du/dx = cos x;
dy/dx = 2u
(dy/dx)(du/dx) = 2(sinx)(cosx)

Int (2sinx cosx) dx = Int (2sinx)d(sinx) = sin^2 x

2007-03-28 13:41:29 補充：
(dy/dx)(du/dx) = 2(sinx)(cosx) = sin2xInt (sin2x) dx = -(1/2)(cos2x) = -(1/2)(2cos^2 x -1)= 1-cos^2x = sin^2 x

2007-03-28 15:24:47 補充：
-(1/2)(2cos^2 x -1) = -cos^2 x (1/2)= -(1-sin2^ x) (1/2)= sin^2x -1 (1/2)= sin^2 x -(1/2)= sin^2 x C-- -(1/2) is constant希望能幫你

d/dx sin^2 x
=2sin x d sin x
= 2 sin x cos x


in (2sin x cos x ) dx
= in (2 sin x ) d (sin x)
= sin^2 x +C

Constant 唔使理啦

OK

2007-03-28 13:37:12 補充：
in (2sin x cos x ) dx= (-1) in (2 cos x ) d (cosx) ***個減一好重要架= -cos^2 x C= sin^2 x C最後果行用左Sin^2 x Cos^2 x = 1Sin^2 x = 1 - Cos^2 x如果用Definite Integral 應該都無問題架