一道高数小题
设∫(0,π)[f(x)+f''(x)]sinxdx=5,f(π)=2,求f'(0).
我的过程:
∫(0,π)f(x)sinxdx+[sinx*f'(x)]|(0,π)-∫(0,π)f'(x)dsinx=5
∫(0,π)f(x)sinxdx-∫(0,π)f'(x)dsinx=5
-∫(0,π)f(x)dcosx-∫(0,π)f'(x)cosxdx=5
-∫(0,π)f(x)dcosx-∫(0,π)cosxdf(x)=5
f(x)cosx|(0,π)=-5
f(0)=-7
我究竟该在哪一步转向呢,毕竟要我求的是f'(0)