∫[-π/2~π/2](sinx/(cosx+x^2))dx
∫[-π/2~π/2](sinx/(cosx+x^2))dx
赶快∫【-π,π】[sinx/(x^2+cosx)]dx和∫【-π/4,-π/3】(sinx+cosx)dx求解
∫x^2 sinx cosx dx ..
∫sinx/(cosx-sin^2x)dx
=∫(0,π/4)(cosx-sinx)dx+∫(π/4,π/2)(sinx-cosx)dx
∫(sinx+cosx)^2 dx
∫(cosx)^2/(cosx-sinx)dx
求∫sinx dx/(sinx+cosx)的积分,x/2-ln|sinx+cosx|+c
∫(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)
求定积分∫(-π/2→π/2)(x|x|+cosx)dx/[1+(sinx)^2]
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