Let $g:\left[0,\dfrac{π}{2}\right]\to \mathbb{R}$ be $$g(x) = \int_{\cos{x}}^{\sin{x}}\sqrt{1-t^2}\,\mathrm{d}t.$$
I need to find the value of $g'(x)$. Can you give me any hints how to do that?
If $G$ is a primitive of $g$, then $g(x)=G(\sin{x})-G(\cos{x})$, but I do not know if it is the right way or what to do after.
On
What is wrong with solving the trivial integral??
$$\int\sqrt{1 - t^2}\ dt = \frac{1}{2} \left(t \sqrt{1-t^2}+\sin ^{-1}(t)\right)$$
Hence just substitute $t = \cos(x)$ and $t = \sin(x)$ and thou hast
$$\frac{1}{2} \left(\sin ^{-1}(\sin (x))+\sin (x) \sqrt{\cos ^2(x)}-\sin ^{-1}(\cos (x))-\sqrt{\sin ^2(x)} \cos (x)\right)$$
Then take the easy derivative which gives you
$$g'(x) = \sin (x) \sqrt{\sin ^2(x)}+\cos (x) \sqrt{\cos ^2(x)}$$
The square roots, due to the chosen range, can be written as the modulus of the sine and cosine respectively.
Hence
$$g'(x) = \sin(x)|\sin(x)| + \cos(x)|\cos(x)|$$
Let $F(x) = \int_0^x \sqrt{1 - t^2} \ dt$. Then $g(x) = F(\sin x) - F(\cos x)$ and so $g'(x) = F'(\sin x) \cos x + F'(\cos x) \sin x$. Noting that $F'(x) = \sqrt{1 - x^2}$ we obtain $$ g'(x) = |\cos x| \cos x + |\sin x| \sin x. $$