Question: Find the derivative of $\int^x_3{\sin^3t}dt$
Since $f(x)$ is a trig polynomial, $f(x)$ is continuous, so I applied FTOC I.
$\frac{d}{dt}(\int^x_3{\sin^3t}dt$)
= $F'(x)$
= $\sin^3x$ (substituted x into t)
Apparently this is not the answer, can someone tell me what I'm doing wrong here?
On
The Fundamental Theorem of Calculus states that if $$g(x) = \int_{a}^{f(x)} h(t)~{\rm d}t$$ where $a$ is any constant, then $$g'(x) = h(f(x)) \cdot f'(x)$$ Using this with the integral, $g(x) = g(x)$, $f(x) = x$, and $h(x) = \sin^3 t$. We may now plug in these values. $$g'(x) = \sin^3(x)$$ So, $$\frac{d}{dx}\Big\lbrack\int_3^x\sin^3(t)dt\Big\rbrack=\sin^3(x)$$
It seems that you are confused about what the problem is actually asking. According to your post, we want to evaluate $$\frac{d}{dx}\Big\lbrack\int_3^x\sin^3(t)dt\Big\rbrack$$ By the fundamental theorem of calculus, we can "replace" the $t$ and $x$, to arrive at the answer $$\sin^3(x)$$
However, if the question was to simply evaluate the integral $$\int_3^x\sin^3(t)dt$$ then we have $$\dfrac{\cos^3\left(x\right)-3\cos\left(x\right)}{3}-\dfrac{\cos^3\left(3\right)-3\cos\left(3\right)}{3}$$
The two procedures above are very different, since taking the derivative of an integral (i.e. applying $\frac{d}{dx}$) is to simply "undo" the integrating process and be left with the original function, whereas integrating the function is to find the antiderivative.