# Finding the derivative of an integral using the Fundamental Theorem of Calculus

How can I find the derivative of $$\int_1^x \sin (x+t) \,dt$$ using the Fundamental Theorem of Calculus? Thanks in advance!

If $$T(x,y)=\int_1^x \ \sin\ (y+t)\ dt$$, then $$T_x=\sin\ (y+x),\ T_y=\int_1^x\ \cos\ (y+t)\ dt$$

Hence $$\frac{d}{dx}\ T(x,x) = T_x+T_y =\sin\ (2x) +\int_1^x \ \cos\ (x+t)\ dt$$

Let f(x) be a continous function. Let F(x) be the antiderivative of that function. The fundamental theorem of calculus tells you that $$\frac{d}{dx}\int_a^b f(x) dx = F'(a) - F'(b) = f(a) - f(b)$$. You have the case however where a or b are functions of x. In that case note that $$\frac{d}{dx}F(g(x)) = g'(x)F'(g(x))$$ by the chain rule. Hence if $$f(x) = sin(x+t)$$, and $$a,b = 1,x$$ then your integral is $$\frac{d}{dx}x * f(x) - f(1) = 1 * F'(g(x)) - sin(x+1) = 2sin(2x) - sin(x-1)$$

Another approach is to use

$$\sin(x+t)=\sin x\cos t+\sin t\cos x$$

$$\sin x\int_1^x\cos t\,dt+\cos x\int_1^x\sin t\,dt$$