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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