# Derivative of integral of $\sin (t^2)$

I'm stuck with the problem

If $F(x)=\int_0^{x^3} \sin t^2 dt$ find $F'(x)$

Now, if the upper interval were $x$, the answer would be $\sin t^2$ (right?). However, the upper interval is $x^3$.

I've thought of just working through the integral and then taking the derivative of the answer, but I don't have a clue how to integrate $\sin t^2$ (not $\sin^2t$). I'm not sure if this is even integrable. If it isn't, is the problem solvable at all?

We can't just put $x^3$ in for $t$, so I'm out of ideas of how to solve this problem.

• Chain rule. If $G(x)=\int_0^x \sin t^2\,dt$, then $F(x)=G(x^3)$ Commented Jan 14, 2014 at 15:31
• I don't think I've ever used the chain rule with integration. So, we don't have to bother integrating $sin t^2$ then? Commented Jan 14, 2014 at 15:35
• Right, you don't have to integrate. You have a function $G(x)=\int_0^x \sin t^2\,dt$. Its derivative is $G'(x)=\sin x^2$. By the chain rule $$[F(x)]'=[G(x^3)]'=G'(x^3)\cdot(x^3)' =(\sin (x^3)^2)\cdot3x^2.$$ Commented Jan 14, 2014 at 15:41

In general, $$\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)dt=f(v(x))v^\prime(x)-f(u(x))u^\prime(x).$$ So, in your case, letting $f(t)=\sin(t^2),$ \begin{align}F^\prime(x)&=f(x^3)(x^3)^\prime-f(0)\cdot 0\\&=\sin\{(x^3)^2\}\cdot 3x^2-0\\&=3x^2\sin(x^6).\end{align}
• so in his case it would be $sin(x^3)*3*x^2$ ? Commented Jan 14, 2014 at 15:36
• NO. $F^\prime(x)=\sin\{(x^3)^2\}\cdot 3x^2.$ Commented Jan 14, 2014 at 15:37
Let $G(t)$ be an antiderivative of $\sin(t^2)$
Then $$\frac{d}{dx} \int_0^{x^3} \sin(t^2) dt = \frac{d}{dx} G(x^3)-G(0) = 3x^2G'(x^3) = 3x^2\sin(x^6)$$