問
f(x) = sinx + |f(x-t)sint dt
f(0) = 0
f'(0) = 0
f(x) = ?

解
f(x) = sinx + A 　とおく。
f''(x) = cosx
f"(x) = -sinx

f(x) = sinx + |f(x-t)sint dt
= sinx +[-f(x-t)cost] + |f'(x-t)cost dt
= sinx - f(x) + [f'(x-t)sint] - |f"(x-t)sint dt
= sinx - f(x) + |sinx sint dt
= sinx - f(x) + sinx|sint dt
= sinx - f(x) + sinx[-cosx]
= sinx - f(x) - sinxcosx - sinx
= - f(x) - sinxcosx
よって、
f(x) = - f(x) - sinxcosx
となるから、
2f(x) = -sinxcosx

f(x) = -sinxcosx/2
~~~~~~~~~~~~~~~~~~
これで合ってますか？