Derivative of $f(x)=\int_{x}^{\sqrt {x^2+1}} \sin (t^2) dt$

Derivative of $$f(x)=\int_{x}^{\sqrt {x^2+1}} \sin (t^2) dt$$

Firstly I wanted to calculate $$\int \sin (t^2) dt$$ and then use $$x$$ and $$\sqrt {x^2+1}$$. But this antiderivative not exist so how can I do this? Is this function at all possible to count?

Let $$f(t)=\sin(t^2)$$.

$$f$$ is continuous at $$\Bbb R$$, thus

$$\displaystyle{F: x\mapsto \int_0^xf(t)dt}$$ is differentiable at $$\Bbb R$$ and for all $$x\in \Bbb R$$,

$$F'(x)=f(x)=\sin(x^2)$$

but

$$G(x)=\int_x^{\sqrt{x^2+1}}f(t)dt=$$ $$F(\sqrt{x^2+1})-F(x)$$

with $$x\mapsto \sqrt{x^2+1}$$ differentiable at $$\Bbb R$$. thus by chain rule, $$G$$ is differentiable at $$\Bbb R$$ and

$$G'(x)=\frac{2x}{2\sqrt{x^2+1}}F'(\sqrt{x^2+1})-F'(x)$$

$$=\frac{x}{\sqrt{x^2+1}}\sin(x^2+1)-\sin(x^2).$$

Hint: Use the Fundamental Theorem of Calculus, which says that if $$I(x) = \int_0^x \sin(t^2)\,{\rm d}t,$$then $$I'(x) = \sin(x^2)$$. Now observe that $$f(x) = I(\sqrt{x^2+1}) - I(x)$$ and use the chain rule.

Use the Lebinitz rule.

$$f'(x) = \sin(x^2+1) \times \frac{2x}{2\sqrt{x^2+1}} - \sin(x^2)$$

Let $$g(x)=\int_0^{x} \sin (t^{2})dt$$. Then $$f(x)=g(\sqrt {1+x^{2}}) -g(x)$$. ALso $$g'(x)=\sin (x^{2})$$. Can you now compute $$g'$$ using Chain rule?