Integral Question- Help would be appreciated! $\int_{1}^{x^2} t^2 \cos(\pi t)$ Just started talking about integration in calculus class.
I'm thrown off by this problem:
Let $$f(x) = \int_{1}^{x^2} t^2 \cos(\pi t)$$
Evaluate $F'(x)$ and $F'(\sqrt{3})$.
By the way, we just started integration so the only technique I know is u-sub.
What throws me off is there being two variables (why is $x^2$ in the upper boundary?)
 A: $$f(x)=\int_{1}^{x^2} t^2 \cos(\pi t)\,\mathrm dt$$
Using Leibniz Integral Rule
 We have

$$\frac{\partial}{\partial x}\int_{a(x)}^{b(x)}p(t,x)\,\mathrm dt=\int_{a(x)}^{b(x)}\frac{\partial p}{\partial x}\,\mathrm dt+p(b(x),x)\frac{\partial b}{\partial x}-p(a(x),x)\frac{\partial a}{\partial x}$$

Now put $$p(t,x)=t^2 \cos(\pi t)$$
So,
$$\begin{align}f'(x)&=\int_{1}^{x^2} 0\,\mathrm dt + (x^2)^2 cos(\pi x^2)\cdot 2x- 1^2\cos(\pi\cdot 1)\cdot0\\
&=[0]_{1}^{x^2}+2x\cdot(x^2)^2 \cos(\pi x^2)-0\\
&=0+2x\cdot(x^2)^2 \cos(\pi x^2)\\
&=2x\cdot(x^2)^2 \cos(\pi x^2)\\
f'(x)&=2x^5\cos(\pi x^2)\\
\end{align}$$
and then

$$f'(\sqrt3)=-18\sqrt3$$

A: The function in the integral is continuous, differentiable and all everywhere, thus it has a primitive function, say $\;G\;$ , and thus
$$F(x):=\int\limits_1^{x^2}t^2\cos\pi t\;dt=G(x^2)-G(1)\implies$$
$$F'(x)=\left(G(x^2)\right)'=2x\,G'(x^2)=2x(\left.t^2\cos\pi t)\right|_{t=x^2}=2x(x^4\cos\pi x^2)=...$$
A: Note that
$${d\over du}\int_a^ u h(t)\ dt= h(u)\qquad\forall u\ .\tag{1}$$
One may view this as a form of the FTC or as an immediate corollary to the definition of $\int_a^b h(t)\>dt$. 
Now we have $$f(x):=g(x^2)\tag{2}$$ where $g$ is defined by
$$g(u):=\int_1^u t^2\cos(\pi t)\>dt\ .$$
Using the principle $(1)$ we get
$$g'(u)=u^2\cos(\pi u)\ .$$
Applying the chain rule to $(2)$ we therefore obtain
$$f'(x)=g'(x^2)\cdot 2x=2x^5\cos(\pi x^2)\ .$$
I can leave the computation of $f\bigl(\sqrt{3}\bigr)$ to you.
A: You have
\begin{equation}
\int_{1}^{x^2}\tau^2\cos\left(\pi\tau\right)d\tau,
\end{equation}
and $\tau$ is simply a "dummy" variable for the integration and could be replaced by anything. Use Leibniz's rule to "integrate" now (actually, it's more taking the derivative and rewriting):
\begin{equation}
F'\left(x\right)=f\left(x\right)=\frac{d}{dx}\int_{g\left(x\right)}^{h\left(x\right)}f\left(x\right)\:dx=f\left(h\left(x\right)\right)h'\left(x\right)-f\left(g\left(x\right)\right)g'\left(x\right),
\end{equation}
which applied to your problem gives us
\begin{align*}
f\left(x\right)=\frac{d}{dx}\int_1^{x^2}\tau^2\cos\left(\pi\tau\right)d\tau=\left(x^2\right)^2\cos\left[\pi\left(x^2\right)\right]\cdot 2x-0=2x^5\cos\left(\pi x^2\right).
\end{align*}
Since you now have that, you may evaluate at $f\left(\sqrt{3}\right)$ since you already have $f\left(x\right)$.
