(19) int tan sqrt (1+{x)^2}cdot dfrac (xdx)(sqrt {1+{x)^2}}-|||-(20) int dfrac (arctan sqrt {x)}(sqrt {x)(1+x)}dx-|||-21 f(1thx)/(t+sx)^2dx;-|||-(22) int dfrac (dx)(sin xcos x)-|||-(23) int dfrac (ln tan x)(cos xsin x)dx-|||-(24)|cosx|xdx-|||-(25) int (cos )^2(omega t+varphi )dt;-|||-circled (19)int sin 2xcos 3xdx;-|||-(27) int cos xcos dfrac (x)(2)dx;-|||-(28)∫sin 5xsin77xdx;-|||-(29)∫tan^2 xsee xdx;-|||-(30) int dfrac (dx)({e)^x+(e)^-x}-|||-(31) int dfrac (1-x)(sqrt {9-4{x)^2}}dx;-|||-(32) int dfrac ({x)^3}(9+{x)^2}dx;-|||-(33) int dfrac (dx)(2{x)^2-1}-|||-34 f(4h)/((x+1(x-2))^2-|||-(35) int dfrac (x)({x)^2-x-2}dx;-|||-36 int dfrac ({x)^2dx}(sqrt {{a)^2-(x)^2}}(agt 0)-|||-17 int dfrac (dx)(xsqrt {{x)^2-1}}-|||-(38) int dfrac (dx)(sqrt {{({x)^2+1)}^x}}-|||-(39) int dfrac (sqrt {{x)^2-9}}(x)dx-|||-40] dfrac (dx)(1+sqrt {2x)}-|||-(41) int dfrac (dx)(1+sqrt {1-{x)^2}}-|||-(42) int dfrac (dx)(x+sqrt {1-{x)^2}}-|||-(43) int dfrac (x-1)({x)^2+2x+3}dx;-|||-(44) int dfrac ({x)^3+1}({({x)^2+1)}^2}dx-|||-前面我们在复合函数求导法则的基础上,得到了换元积分法.现在我-|||-两个函数乘积的求导法则,来推得另一个求积分的基本方法一 爸想-|||-设函数 u=u(x) 及 v=v(x) 具有连续导数,则两个函数乘积的导数化-|||-(uv)'=u'w+uv'-|||-(uv)'=u'w+uv'-|||-移项,得-|||-'=((uv))^t-(u)^vv.-|||-对这个等式两边求不定积分,得-|||-f uv`dx=uv- ∫u`vdx