# I'm confuse with this, please explain

#### kt chua

##### New member
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.

Last edited:

#### Subhotosh Khan

##### Super Moderator
Staff member
Given

f'[(sinx)^2]=(cosx)^2
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

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

f(y)=y-(y^2)/2+c
Couldn't spot where is the mistake.
Please post the EXACT problem as it was given to you.

• topsquark

#### HallsofIvy

##### Elite Member
When you went from "$$\displaystyle f'(sin^2(x))= cos^2(x)$$" to "$$\displaystyle f(sin^2(x))= \frac{cos(x)sin(x)+ x}{2}+ c$$" you integrated the right side or the equation with respect to x but you integrated the left side of the equation with respect to $$\displaystyle sin^2(x)$$.

• Jomo and topsquark

#### kt chua

##### New member
When you went from "$$\displaystyle f'(sin^2(x))= cos^2(x)$$" to "$$\displaystyle f(sin^2(x))= \frac{cos(x)sin(x)+ x}{2}+ c$$" you integrated the right side or the equation with respect to x but you integrated the left side of the equation with respect to $$\displaystyle sin^2(x)$$.
Thank for explanation. This is the part I not clear. I tried with 2 example below
ie

f(x)=1-x^2 and f(sinx)=1-(sinx)^2
f'(x)=-2x f'(sinx)=-2sinxcosx
if x=sinx
f'(sinx)=-2sinx

From this example both f'(sinx) yield different result. Is the notation suppose to be the same in mathematics?

#### Jomo

##### Elite Member
Wait a minute! If f(x) = 1-x^2, then f'(x) = -2x. So f'(sin(x)) = -2sinx so 2xf'(sinx) = 2x*(-2sinx)= -4xsinx NOT -2sinxcosx

I think this is you problem. You think f'(sinx) means to take the derivative of sinx which is cosx. That is NOT correct. f'(sinx) means to evaluate the function, whose name happens to be f', at sinx

Again, if f(x) = 1-x^2, then f'(x) = -2x, so f'(sinx) = -2sinx.

Just for the record, using your understanding of f', you should always say that f'(x) = 1 (ALWAYS), because the derivative of x is 1. But YOU know this is not true, so why use this method for sinx?

#### HallsofIvy

##### Elite Member
$$\displaystyle f'(sin^2(x))= \frac{d(sin^2(x))}{dx}$$.. If you let $$\displaystyle y= sin^2(x)$$ then you have $$\displaystyle \frac{df(y)}{dx}$$ NOT $$\displaystyle \frac{df(y)}{dy}$$ which is what you are trying to say.