Find Critical Numbers of sin^2 x + cos x Function

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Homework Statement


Find any critical numbers of the function


Homework Equations


sin^2 x + cos x


The Attempt at a Solution



I actually have a sort of silly question. Woud sin^2 in the equation be solved using the differeniation rule of d/dx[sin x] = cos x or d/dx [sin u] = (cos u)d/dx u?
 
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d/dx (sin^2 x)= d/dx (sin x * sin x)
= d/dx (sin x) * sin x + sin x * d/dx (sin x)
=2 d/dx (sin x) sin x
=2cos x sin x
 
So I follow that thinking, but that would be the same as d/dx[sin u] right?
 
No- I think you're referring to a case like this:

A=d/dx sin(f(x))
substitue u=f(x)

A=du/dx * d/du sin(u)
=[d/dx f(x) ] cos u
=[d/dx f(x) ] cos(f(x))

Remember, sin^2(x) does not equal sin (sin (x)) or sin (x^2)
rather sin^2(x)= sin (x) * sin (x)
 
Thank you.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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