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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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    True true, by definition $Si'(x) = \frac{sinx}{x}$!2011-12-18

1 Answers 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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    Thanks this site rocks! After-all I might have a chance passing my finals.2011-12-17