How do you find the derivative of the integral $\sin(\ln x)$ $$\frac{d}{dx} \;\left[ \int_{a}^{x^2}\sin(\ln(z))\;dz\right]$$
I'm not sure if I'd have to do the chain rule on the natural logarithm and them $x^2$, or if there is no chain rule at all. Any help would be appreciated, thank you!
 A: We don't actually need to solve the integral. Let $G(x)=\int \sin(\ln z)dz$. Then
$$\frac{d}{dx}\int\limits_a^{x^2} \sin(\ln z)dz=\frac{d}{dx}(G(x^2)-G(a))$$
$$=2x\cdot G'(x^2)-0$$
$$=2x\sin(\ln(x^2))$$
A: First we find $\int \sin(\ln(x))\, dx$. We integrate by parts two times.
$$I=\int \sin(\ln x)\, dx =\sin(\ln x)\cdot x-\int \frac{\cos (\ln x)}{x}\cdot x\, dx$$
$$=\sin(\ln x)\cdot x-\int \cos (\ln x)\, dx$$
$$=\sin (\ln x)\cdot x-\left(\cos(\ln x)\cdot x-\int\frac{-\sin (\ln x)}{x}\cdot x\, dx\right)$$
$$=\sin(\ln x)\cdot x-\cos (\ln x)\cdot x-I$$
Solve for $I$:
$$I=\frac{x(\sin(\ln x)-\cos (\ln x))}{2}+C$$
Assuming $a,x>0$, we get (since $\sin(\ln x)$ is continuous if $x>0$): $$\frac{d}{dx} \;\left[ \int_{a}^{x^2}\sin(\ln(z))\;dz\right]$$
$$=\frac{d}{dx}\left(\frac{x^2(\sin(2\ln x)-\cos(2\ln x))}{2}-\frac{a^2(\sin(2\ln a)-\cos(2\ln a))}{2}\right)$$
The derivative of the constant is $0$, so:
$$=\frac{1}{2}\cdot \frac{d}{dx}\left(x^2(\sin(2\ln x)-\cos(2\ln x))\right)$$
$$=\frac{1}{2}\left(2x(\sin(2\ln x)-\cos(2\ln x))+x^2\left(\frac{2\cos(2\ln x)}{x}-\frac{-2\sin(2\ln x)}{x}\right)\right)$$
$$=2x\sin(2\ln x)$$
A: You should just use the chain rule in conjunction with the first fundamental theorem of calculus.
$$\eqalign{
  & \left\{ \matrix{
  h(x) = \int_a^x {f(t)dt}  \hfill \cr 
  h'(x) = f(x) \hfill \cr}  \right.  \cr 
  & {\left( {h \circ g} \right)^\prime } = \left( {h' \circ g} \right)g' \cr} $$
in your question we have
$$\eqalign{
  & h(x) = \int_a^x {f(t)dt}   \cr 
  & f(t) = \sin (\ln (t))  \cr 
  & g(x) = {x^2} \cr} $$
and hence we have
$${\left( {\int_a^{{x^2}} {\sin (\ln (t))dt} } \right)^\prime } = \sin (\ln ({x^2}))2x$$
