why the answers are difference when using FTC1 and integrate then take derivatives? for example,
the derivatives of $\int_1^x\cos4(x+t)\,\mathrm dt$
using FTC1, the answer is $2\cos(8x)$
but integrate it and then take derivatives, the answer is $2\cos(8x) - \cos(4(x+1))$. Why? 
Thanks
Seems I can rewrite the equation using trig identity.... 
$cos4(x+t) = cos(4x+4t) = cos(4x)cos(4t) - sin(4x)sin(4t) $
then I can apply FTC1 directly, 
$ \int_1^x cos4(x+t)dt$ = cos(4x)$\int_1^x cos(4t)dt - sin(4x)$ $\int_1^x sin(4t)dt$ $ = 2cos(8x) $
but $2cos(8x) $ looks different to 2cos(8x)−cos(4(x+1))
 A: Since we have to differentiate with respect to a variable which appears in the integral itself AND in the limit of the integral, the calculus must respect the rules of differentiation for both.
To make it clear, separate both.
$$\text{Let :} \qquad f(x,y)=\int_1^y\cos4(x+t)\,\mathrm dt$$
$$\frac{\partial f}{\partial y}=\cos4(x+y) \tag 1$$
$\frac{\partial f}{\partial x}=\int_1^y\frac{\mathrm d}{\mathrm d x}\cos4(x+t)\,\mathrm dt = -4\int_1^y\sin4(x+t)\,\mathrm dt =  \left[\cos4(x+t)\right]_{t=1}^{t=y} =\cos4(x+y)-\cos4(x+1) $
$$\frac{\partial f}{\partial x}=\cos4(x+y)-\cos4(x+1) \tag 2$$
Total differential :
$$\mathrm d f(x,y)=\frac{\partial f}{\partial x}\mathrm d x +\frac{\partial f}{\partial y}\mathrm d y = \big(\cos4(x+y)-\cos4(x+1)\big) \mathrm d x +\cos4(x+y) \mathrm d y$$
Case of $\quad y=x$ :
$$\mathrm d f(x,x)= \big(\cos4(x+x)-\cos4(x+1)\big) \mathrm d x +\cos4(x+x) \mathrm d x$$
$$\mathrm d f(x,x)= \big(2\cos8x-\cos4(x+1)\big) \mathrm d x $$
$$\frac{\mathrm d}{\mathrm d x}\int_1^x\cos4(x+t)\,\mathrm dt=2\cos8x-\cos4(x+1)$$
Of course, people familiar with this kind of problem doesn't need such a tiresome detailed calculus. They straightforward apply the rules. 
