2 2dt=(2+-)d [2+2n(t-1) (6+2ln2)-(4+2lnl)=2+2ln2
− = + − = − 9 4 3 2 3 2 )d 1 2 2 d (2 1 1 d 1 1 t t t t t x x 3 2 = [2t + 2ln(t −1)] = (6 + 2ln 2) − (4 + 2ln 1) = 2 + 2ln 2
例2求 a'-xd 解设x= a t,dx= a cos tdt,￥1x=O时,t=0, x=时,t=-,所以 2 xdx=2 acost (acost dt=a2 cos tdz a (2(1+cos 2t dt C几 T (t+-sin 2t 一20 +0) 4
例2 求 解 设 − a a x x 0 2 2 d . x a t x a t t x t = = = = sin ,d cos d , 0 , 0, 当 时 π , , 2 x a t = = 时 所以 π π 2 2 2 2 2 2 0 0 0 d cos ( cos )d cos d a a x x a t a t t a t t − = = 2 π 2 0 (1 cos2 )d 2 a = + t t π 2 2 2 2 0 1 π π ( sin 2 ) ( 0) . 2 2 2 2 4 a a a = + = + = t t
该定积分的几何意义是半径为a,以原点为圆心的 员在第一象限内的面积,即该圆面积的四分之 例3求 sin x cos rdx 解法1 sinx cos xax sin xa(sinx 0 sin x
该定积分的几何意义是半径为 以原点为圆心的 圆在第一象限内的面积，即该圆面积的四分之一. a, 2 0 sin cos d . x x x x 解法1 π 2 2 0 0 sin cos d sin d(sin ) x x x x x x = . 2 1 sin 2 1 2 0 2 = = x x 例3 求