Find the integral $\frac{d }{d x}\int_{\sin x}^{\cos x}\cos ( \pi t^2)\, d t $ Find the integral $$\frac{d}{dx}\int_{\sin x}^{\cos x}\cos ( \pi t^2) \, dt. $$


Am I right that it is $$\frac{d}{dx}\cos x \cdot  \cos (\pi \cdot  \cos^2x) - \frac{d}{dx} \sin x \cdot  \cos (\pi \cdot  \sin^2x))?$$
$$\int_a^b \cos ( \pi t^2))\,\mathrm d t = \frac{1}{\sqrt{2}}\left[C(\sqrt{2} b)-C(\sqrt{2}a)\right].$$  This is a Fresnel integral.

$$\int_{\sin x}^{\cos x}\cos ( \pi t^2))\,\mathrm d t  =\frac{1}{\sqrt{2}}\left[C(\sqrt{2} \cos x)-C(\sqrt{2}\sin x)\right].$$

$$\frac{\mathrm{d} }{\mathrm{d} x}\int_{\sin x}^{\cos x}\cos ( \pi t^2))\,\mathrm d t =-\sin x \, C^\prime(\sqrt{2} \cos x)- \cos x \, C^\prime(\sqrt{2}\sin x). $$

$$= -\sin x \cdot  \cos (\pi \, \cos^2x) -  \cos x \cdot  \cos (\pi \, \sin^2x))$$ 

or just use the fundamental theorem of calculus, as you did.
$$\int_a^b \cos ( \pi t^2))\,\mathrm d t = \frac{1}{\sqrt{2}}\left[C(\sqrt{2} b)-C(\sqrt{2}a)\right].$$  This is a Fresnel integral.

$$\int_{\sin x}^{\cos x}\cos ( \pi t^2))\,\mathrm d t  =\frac{1}{\sqrt{2}}\left[C(\sqrt{2} \cos x)-C(\sqrt{2}\sin x)\right].$$

$$\frac{\mathrm{d} }{\mathrm{d} x}\int_{\sin x}^{\cos x}\cos ( \pi t^2))\,\mathrm d t =-\sin x \, C^\prime(\sqrt{2} \cos x)- \cos x \, C^\prime(\sqrt{2}\sin x). $$

$$= -\sin x \cdot  \cos (\pi \, \cos^2x) -  \cos x \cdot  \cos (\pi \, \sin^2x))$$ 

or just use the fundamental theorem of calculus, as you did.