=∫(0,π/4)(cosx-sinx)dx+∫(π/4,π/2)(sinx-cosx)dx

=∫(0,π/4)(cosx-sinx)dx+∫(π/4,π/2)(sinx-cosx)dx ∫π0(sinx+cosx)dx ∫(0,π)|sinx-cosx|dx 赶快∫【-π,π】[sinx/(x^2+cosx)]dx和∫【-π/4,-π/3】(sinx+cosx)dx求解 ∫f（sinx,cosx）dx=∫f（cosx,sinx）dx上下限是[0,π/2] ∫(0,10π)[(sinx)^3]/[2(sinx)^2+(cosx)^4]dx 证明：积分符号sinx/(sinx+cosx)dx=积分符号cosx/(sinx+cosx)dx在[0,π/2]相等 加 ∫(0,π/2)(-sinx+cosx)/(sinx+cosx)dx 请用换元法求出定积分 ∫【0到π/2】(sinx^10-cosx^10)dx/(5-sinx-cosx) ∫cosx / (cosx+sinx)dx ∫dx/（sinx+cosx) ∫(sinx-cosx)dx