赶快∫【-π,π】[sinx/(x^2+cosx)]dx和∫【-π/4,-π/3】(sinx+cosx)dx求解
赶快∫【-π,π】[sinx/(x^2+cosx)]dx和∫【-π/4,-π/3】(sinx+cosx)dx求解
=∫(0,π/4)(cosx-sinx)dx+∫(π/4,π/2)(sinx-cosx)dx
∫[-π/2~π/2](sinx/(cosx+x^2))dx
∫π0(sinx+cosx)dx
∫(0,π)|sinx-cosx|dx
∫(0,π)x(e^sinx)|cosx|dx
∫(0,10π)[(sinx)^3]/[2(sinx)^2+(cosx)^4]dx
∫(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)
∫f(sinx,cosx)dx=∫f(cosx,sinx)dx上下限是[0,π/2]
∫dx/(2+sinx) 和∫dx/(3+cosx)
∫cosx / (cosx+sinx)dx