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I would appreciate if somebody could help me with the following problem:

Q: Find $f(x)$ ($f'(x)$: conti-function , $x \in\mathbb{R}$) $$f(x)=\sin ^2x+\int_{0}^{x}tf(t)dt$$

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Here is what google returns for the query "conti function". – 1015 Jun 1 '13 at 3:45

$f '(x)-xf(x)=2\sin(x)\cos(x)$ $\to$ $$f(x)=e^{\left(-\dfrac{x^2}{2}\right)} \left(\int_0^x e^{\left(-\dfrac{t^2}{2}\right)}\cdot2\sin(t)\cos(t)dt+c\right)$$

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Please check my edit and see if messed something up. Sorry if I did! – Amzoti Jun 1 '13 at 3:39
That ought to be $e^{-\frac{x^2}{2}}$ in the integral, I believe. – Zen Jun 1 '13 at 3:41
@Amzoti:thanks u so much – Maisam Hedyelloo Jun 1 '13 at 3:52
@MaisamHedyelloo You can evaluate the value of constant $c$. Put $x =0$ in the equation to get the initial condition $f(0) = 0$. I hope this will give $c = 0$. If it is correct you may edit your answer. – srijan Jun 1 '13 at 4:33

Differentiate both sides wrt x to get $$f'(x)=2\sin(x)\cos(x)+xf(x)$$ which is a linear ODE in f(x). Then use the integrating factor method. You'll get f(x) down to an integral in terms of x, but it's not going to be an elementary antiderivative.

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I'm very silly, I apologize for the earlier comment – DanZimm Jun 1 '13 at 6:34

It's basically a fundamental theorem of calculus problem. I was thinking your only confusion would probably be how to get rid of $t$. Remember the fundamental theorem of calculus says if $F(x)=\int_0^t f(t)dt$, then we have $F'(x)=f(x)$. I think that's probably the only trick you need to use. The rest of this question is just standard ODE solving.

Please don't select my answer. It's just meant to complement other answers.

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