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

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3 Answers 3

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

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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 $\frac{d}{dx}x * f(x) - f(1) = 1 * F'(g(x)) - sin(x+1) = 2sin(2x) - sin(x-1)$

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Another approach is to use

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

so your function is

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

Now differentiate each term using the product rule and the fundamental theorem.

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