What is this question asking? (integrals, derivatives)

The question is:

Let $G(x) = \int_{\cos(2x)}^{1/2} \arcsin(t) dt$. Find all $x \in [0,\pi/2]$ such that $\frac{d}{dx} G(x) = 0$.

Is it asking "for what values of x between 0 to $\pi/2$ is the derivative 0?" And if it is how would I proceed.

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Hint: use the Fundamental theorem of Calculus to write down $G'(x)$, then solve for $G'(x) = 0$. – user44441 Jan 29 '13 at 1:56

@amWhy noted the way of solving theoretically, but: $$G(x)=-\int_{0.5}^{\cos(2x)}\arcsin(t)dt$$ and so $$G'(x)=2\arcsin(\cos(2x))\sin(2x)$$ and now, we have: $$\arcsin(\cos(2x))=0~~~\text{or}~~~\sin(2x)=0$$ If $$\arcsin(\cos(2x))=0\Longrightarrow\cos(2x)=\sin(0)=0$$ and then $$\cos(2x)=0\longrightarrow2x=2k\pi\pm\pi/2,~~~k\in\Bbb Z$$ Now pick up the values you are wanted to chose. And if $\sin(2x)=0$ then $$2x=k\pi,~~~k\in\Bbb Z$$. I think the common solutions which lie in the interval can be easily chosen.

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nice work! +1 :-) – amWhy Jan 29 '13 at 3:48
If $F(x)=\int_{a(x)}^{b(x)} g(t)dt$ then $F'(x)=g(b(x)) b'(x)-g(a(x))a'(x)$. In the above $G'(x)$ is missing derivative of $\cos(2x)$. – Maesumi Jan 29 '13 at 4:29
@Maesumi: Yes, you were right. I fixed it. Thanks for noting me that. +1000 – S. Snape Jan 29 '13 at 4:43
Nice graphic! +1 if I could upvote again! – amWhy Jan 29 '13 at 4:55
@amWhy: It is kind of you. I did it maybe the OP gets the solutions in the interval better. – S. Snape Jan 29 '13 at 5:35

Hint: Make use of the Fundamental Theorem of Calculus. Given $G(x)$ is the antiderivative of $g(t)\,dt$, when evaluated, is a function of $x$, whose derivative is then what in terms of the integrand? (It is integrand, essentially, but expressed as function $g(x)$ of $x$ and no longer a function of $t$, given the bounds of integration.)

Then, yes, find the values of $x$ for which the derivative of the integral $g(x)= \frac {d}{dx}\left(G(x)\right)\,=\,0$.

Note that in evaluating your integral, and viewing it in terms of the Fundamental Theorem of Calculus: $$G(x)=\int_{\cos(2x)}^{1/2} \arcsin(t)\,dt = -\int_{0.5}^{\cos(2x)}\arcsin(t)dt$$

so here, "$g(x)$" is really a function of $\cos(2x): g(\cos(2x))$, and so $$\frac{d}{dx}\,G(x)=2\arcsin(\cos(2x))\sin(2x)\tag{\frac{d}{dx}G(x)}$$

Now, $\frac{d}{dx}G(x)$ evaluates to $0$ when either $$\arcsin(\cos(2x)) = 0\tag {1}$$ or when $$\sin(2x) = 0\tag{2}$$

Can you take it from here?

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I did some handy jobs for the OP. + – S. Snape Jan 29 '13 at 3:45
This answer does not take into account the cos term. – Giuseppe Negro Jan 29 '13 at 3:50