∫（xcosx-5sinx+2）dx
=∫xcosxdx-5∫sinxdx+2x+C
=∫xd(sinx)+5cosx+2x+C
=x*sinx-∫sinxdx+5cosx+2x+C
=xsinx+cosx+5cosx+2x+C
=xsinx+6cosx+2x+C
代入上下限得,
∫（xcosx-5sinx+2）dx=4a
∫上a下-a（xcosx-5sinx+2）dx的值
1个回答
相关问题
-
∫2xcosx^2dx = ∫sinx^2dx^2 吗?
-
∫(xcosx-sinx)dx/(x-sinx)^2 求不定积分
-
求解一道不定积分题∫（xcosx-sinx）/x^2dx
-
∫xcosx/(sinx)^2dx分部积分法咋做?
-
求不定积分∫sinx-xcosx/cosx+xsinx dx
-
证明∫(上π,下0)xf(sinx)dx=π/2∫(上π,下0)f(sinx)dx
-
∫(xcosx+|sinx|)dx上限是π下限是-π的定积分
-
∫xcosx/2dx积分求解
-
∫(xcosx)^2dx的积分怎么做?
-
若f(x)为[-a,a]上连续偶函数,求证:∫(-a下a上）f(x)dx=2∫(0下a上）f(x)dx.