Given an equation
f'[(sinx)^2]=(cosx)^2
Find f(x)
The solution is
By substituting y=(sinx)^2 and using trigonometry identity, this equation is written as
(sinx)^2+(cosx)^2=1
f'(y)=1-y
f(y)=y-(y^2)/2+C
hence
f(x)=x-(x^2)/2+c
However if I didn't do the substitution earlier while directly solve the integration for (cosx)^2, will yield
f[(sinx)^2]=(cosxsinx+x)/2+c
and now try to substitute y=(sinx)^2
I fail to obtain the f(y) like above equation
f(y)=y-(y^2)/2+c
Couldn't spot where is the mistake.
f'[(sinx)^2]=(cosx)^2
Find f(x)
The solution is
By substituting y=(sinx)^2 and using trigonometry identity, this equation is written as
(sinx)^2+(cosx)^2=1
f'(y)=1-y
f(y)=y-(y^2)/2+C
hence
f(x)=x-(x^2)/2+c
However if I didn't do the substitution earlier while directly solve the integration for (cosx)^2, will yield
f[(sinx)^2]=(cosxsinx+x)/2+c
and now try to substitute y=(sinx)^2
I fail to obtain the f(y) like above equation
f(y)=y-(y^2)/2+c
Couldn't spot where is the mistake.
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