Calculating Integral Substituting and Showing Convergence in x\rightarrow0

[wel]

Consider the integral
\begin{equation}
I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt
\end{equation}
show that
\begin{equation}
I(x)= \frac{2x}{\pi} +O(x^{3})
\end{equation}
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.
please help me.

 

[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).

 

[Saitama]

Consider the integral
\begin{equation}
I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt
\end{equation}
show that
\begin{equation}
I(x)= \frac{2x}{\pi} +O(x^{3})
\end{equation}
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.
please help me.

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 $I'(0)$?

 

[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?

 

[SammyS]

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?

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