Question

求解这题不定积分

Answer 1

∫ (-π/2->π/2) (xsinx + cosx) dx
=2∫ (0->π/2) (xsinx + cosx) dx
=2{∫ (0->π/2) xsinx dx + [sinx]| (0->π/2) }
=2{ -∫ (0->π/2) xdcosx + 1 }
=2{ -[xcosx]|(0->π/2) + ∫ (0->π/2) cosx dx + 1 }
=2{ [sinx]|(0->π/2) + 1 }
=4

追问

思路是

追答

f(x)=xsinx + cosx
f(-x) = f(x)
∫ (-π/2->π/2) (xsinx + cosx) dx
=2∫ (0->π/2) (xsinx + cosx) dx
=2∫ (0->π/2) xsinx dx +2∫ (0->π/2) cosx dx
分部积分
=-2∫ (0->π/2) xdcosx +2
=-2[xcosx]|(0->π/2) + 2∫ (0->π/2) cosx dx + 2
=2∫ (0->π/2) cosx dx + 2

=2 [sinx]|(0->π/2) +2
=2+2
=4