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I need to find $S'(x)$ where:

$$S(x) = \int_1^{x^2}\frac{\sin{t}}{t} dt$$

My first thought was to solve the definite integral first getting a function that only depends on $x$ and then derivate that.

However, trying to find the primitive of $\frac{\sin t}{t}$ seems impossible. Wolfram Alpha gives me $\int\frac{\sin{t}}{t} dt=\mathrm{Si}(t)+C$ where $\mathrm{Si}(x)$ is the sine integral. So I don't get an expression that I can derivate but a definition.

On the other side this problem should be easier than that. Let me explain:

If $$F(x) = \int{x^2} dx$$ then $$F'(x) = x^2 +C$$

I should be able to apply that to $S(t)$ but how?

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    $\begingroup$ The sine integral is fine. Since $\mathrm{Si}(x)$ is defined as the integral of $\dfrac{\sin\,x}{x}$, it stands to reason that $\mathrm{Si}^\prime(x)=\dfrac{\sin\,x}{x}$. $\endgroup$ Commented Dec 18, 2011 at 3:48
  • $\begingroup$ Yeap, but my problem was how do you calculate the value of $Si(x^2)-Si(1)$ to find out the definite integral? $\endgroup$
    – dimme
    Commented Dec 18, 2011 at 12:43
  • $\begingroup$ The chain rule remains applicable. $\mathrm{Si}(1)$ is constant, and $\dfrac{\mathrm d}{\mathrm dx}\mathrm{Si}(x^2)=\mathrm{Si}^\prime(x^2)\dfrac{\mathrm d}{\mathrm dx}x^2$... $\endgroup$ Commented Dec 18, 2011 at 12:47
  • $\begingroup$ True true, by definition $Si'(x) = \frac{sinx}{x}$! $\endgroup$
    – dimme
    Commented Dec 18, 2011 at 12:58

1 Answer 1

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Use the Fundamental Theorem of Calculus: $${d\over dx}\int_a^x f(t)\, dt=f(x)$$ and the chain rule.

So if $F(x)=\int_1^{x } {\sin t\over t}\,dt$, then $$S(x)=F(x^2)$$ and $${d\over dx}S (x)=F'(x^2)\cdot (x^2)'.$$

You have: $${d\over dx}\underbrace{\int_1^{x^2} {\sin t\over t}\,dt}_{F(x^2)}= \underbrace{\sin(x^2)\over x^2}_{F'(x^2)}\cdot (x^2)'=2x{\sin(x^2)\over x^2}.$$

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    $\begingroup$ @Dylan Moreland Thanks Dylan; I can never remember if it's the first or the second (or if this terminology is even consistent) :) $\endgroup$ Commented Dec 17, 2011 at 20:38
  • $\begingroup$ Same here. On Wikipedia and in Stewart (what I TA from) it's the "first", but why? Rudin doesn't call it anything! $\endgroup$ Commented Dec 17, 2011 at 20:43
  • $\begingroup$ I just call it the derivative of integral form of the FTOC. $\endgroup$ Commented Dec 17, 2011 at 21:59
  • $\begingroup$ Thanks this site rocks! After-all I might have a chance passing my finals. $\endgroup$
    – dimme
    Commented Dec 17, 2011 at 22:55

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