例12)计章r4nx dx x 闻原式=+ 4 Inx √x nx 2、xd) x 2√x +(2x Inx dx) x -(2NxInx-4vx)+(2vxInx-4vx 202/2/l1 =6ln2-2
2021/2/1 11 + − = 4 1 1 ln ln 4 1 dx x x dx x x 原式 4 4 1 ln [ 12] dx x x 例 计算 ) 2 (2 ln 1 1 4 1 4 1 | = − − dx x x x x | | 4 1 1 4 1 = −(2 x ln x − 4 x ) + (2 x ln x − 4 x ) = 6ln 2 − 2 ) 2 (2 ln 4 1 4 1| + − dx x x x x [解]
例3]计算x(1-x)k x)"x+ (1-x)"l M+1 M+1 n+2 (n+1)(n+2) (n+1)(n+2)0 n+3 (n+l(+2)(n+3) (n+1)m+2(n+3) 202/2/1
2021/2/1 12 x x dx n − 1 0 2 [例13] 计算 (1 ) x x dx n x x n x x dx n n n + + − + − + + = − − 1 0 1 1 0 1 2 1 0 2 (1 ) 1 2 (1 ) 1 1 (1 ) | x dx n n x x n n n n + + − + + + − + + = − 1 0 2 1 0 2 (1 ) ( 1)( 2) 2 (1 ) ( 1)( 2) 2 | [解] ( 1)( 2)( 3) 2 (1 ) ( 1)( 2)( 3) 2 | 1 0 3 + + + − = + + + = − + n n n x n n n n
丌 一例4计算:n=smd n∈ N) 丌 丌 丌 21dx 71=|2 sin xdx=-(0sx2=1 , =2 sin-xd (-coS x =(-C0m吹r 丌 cos asin 丌 H-4 SIn r cos x SInx·l-sInx x →1n=(n-1)/n2-(n-1)n 202/2/1
2021/2/1 13 [ 14] sin ( ) 2 0 I xdx n N n n = 例 计算： 2 1 2 0 0 = = I dx sin cos | 1 2 0 2 0 1 = = − = I xdx x [解] = − 2 − 0 1 sin ( cos ) I xd x n n n x xdx n = − 2 − 0 2 2 ( 1) sin cos − − = − − − 2 0 2 1 0 1 ( cos )sin | ( cos ) (sin ) x x x d x n n n x x dx n = − − 2 − 0 2 2 ( 1) sin (1 sin ) n n n I (n 1)I (n 1)I = − −2 − −
2(n≥2) 当n=2时,得到 d x= k-1)!丌 SIn x (2k)!2 当n=2k-1时,得到 2-=2sn1r= 2k-2) 202/2/l1
2021/2/1 14 当n = 2k时,得到 (2 )!! 2 (2 1)!! sin 2 0 2 2 − = = k k I x dx k k ( 2) 1 2 − = I − n n n In n 当n = 2k − 1时,得到 1 (2 1)!! (2 2)!! sin 2 0 2 1 2 1 − − = = − − k k I x dx k k