# Integral substituting

1. Apr 19, 2014

### wel

Consider the integral

I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt

show that

I(x)= \frac{2x}{\pi} +O(x^{3})

as $x\rightarrow0$.

=> $sin(x.sint)= x.sint - \frac{(x-sint)^3}{3!}+...$

and integrate term by term should give
$- x.sint - \frac{1}{12}(cos3t-9cost)+...$
when substituting $t=\pi$ and $t=0$ something else comes up.

2. Apr 19, 2014

### CAF123

As x → 0, the linear approximation of sin is enough and indeed the question wants you to ignore further terms in the expansion (the next term being cubic).

3. Apr 19, 2014

### Saitama

Hi wel!

As suggested in your previous thread, solve it in the following way:
$$I(x)=I(0)+I'(0)x+I''(0)\frac{x^2}{2!}+O(x^3)$$
Obviously $I(0)=0$.

Can you find $I'(x)$ and then $I'(0)$?

4. Apr 19, 2014

### wel

$sin(x\cdot sint) = x\cdot sint - \dfrac{(x\cdot sint)^3}{3!} + ...$,
then I guess I have to integrate term by term.
$I(0)=0$
integrate $I(x)= x sint$ at $t =\pi$ and $t=0$ gives the answer but I really don't know why and how?

5. Apr 19, 2014

### SammyS

Staff Emeritus
Which part(s) of this procedure don't you understand ?