Integration by expansion

Question

Consider the integral \begin{equation} I(x)= \frac{1}{\pi} \int^{\pi}_{0} \sin(x\sin t) \,dt \end{equation} show that \begin{equation} I(x)= \frac{2x}{\pi} +O(x^{3}) \end{equation} as $x\rightarrow0$.

=> I Have used the expansion of McLaurin series of $I(x)$ but did not work. please help me.

Answer 1

Just appeal to the Taylor expansion of $I(x)$ directly; clearly $I(0)=0$. Now, $$ I'(x) = \frac{1}{\pi}\int_0^\pi \cos(x\sin{t})\sin{t}\,dt. $$ So, $$ I'(0) = \frac{1}{\pi} \int_0^\pi \sin{t}\,dt = \frac{2}{\pi}. $$ Also, $$ I''(x) = -\frac{1}{\pi}\int_0^\pi \sin(x\sin{t})\sin^2{t}\,dt. $$ So, $I''(0)=0$.

Therefore, $$ I(x) = I(0) + I'(0)x + \frac{I''(0)}{2}x^2 + O(x^3) = \frac{2x}{\pi} + O(x^3). $$

Answer 2

$sin(x\cdot sint) = x\cdot sint - \dfrac{(x\cdot sint)^3}{3!} + ...$, and integrate term by term should give the answer.