习题5?3
1? 计算下列定积分?
(1)??sin(x?)dx?
32 解
?? ??cos4?2?11?cos???0? 3322?21???sin(x?)dx??cos(x?)33???2
(2)??2 解
dx11?2???2(11?5x)35??2(11?5x)?(11?5x)31dx?
1?2??1151? ?16?2??1?2?1010512
(3)?02sin?cos3?d??
解
??20sin?cos3?d????2scos3?dsin?0?
?1 ??cos4?2??1cos4??1cos40?1?
044244 (4)?0(1?sin3?)d?? 解
?
?0?0?(1?sin3?)d???0d???0sin2?dcos????0????0(1?cos2?)dcos??
1 ???(cos??cos3?)3 (5)??2cos2udu?
64????
3
? 解
???2cos261?1udu???2(1?cos2u)du?u262??261?sin2u4??2
61??1??3 ?(?)?(sin??sin)???
2264368
(6)?0 解
22?x2dx?
?02?令x?2sint?22cost?2costdt?2(1?cos2t)dt2?x2dx?0?0?201 ?(t?sin2t)2??2?
(7)?? 解
228?2y2dy?
令y?2sinx2?4?2cosx?2cosxdx?4??228?2y2dy?2???224?y2dy?
?22?4(1?cos2x)dx?2??412(x?sin2y)2?4??4?2(??2)?
(8)?1121?x2x2dx?
解
?1121?x2x2?令x?sint?cost122(dx?costdt???sin2t??sin2t?1)dt?(?cott?t)44??24?1??4?
a (9)?0x2a2?x2dx? 解
?0xa2令x?asint?a4222a?xdx?0asint?acost?acostdt?4224a?02(1?cos4t)dt?8t?02sin?22tdt
a4 ?8??20a4?sin4t32?20a4?? ?16
(10)?13dxx21?xdx2?
解
?13令x?tant2x?21?x??tan2t?sect?sec34?12tdt
1dt?? ???2sint4sint3cost??34?2?23? 3
(11)??1 解
1xdx5?4x?
1??141xdx令5?4x?u111132(5?u)du??(5u?u)?38835?4x1?? 36
(12)
?11?dxx?
解
?11?14dx令x?ux211?11?u?2udu?2?1(1?1?u)du?2(u?ln|1?u|)2212?2(1?ln)?
3
dx1?x?1dx1?x?1 (13)?34?
解
?
134令1?x?u?110112?(?2u)du?20(1?)du?2(u?ln|u?1|)210u?12u?1??1?2ln2?
(14)?02axdx3a?x22?
解
?02axdx3a?x22??12a1d(3a2?x2)??3a2?x2?023a2?x22a0?a(3?1)?
(15)?0te 解
1?t22dt?
?t2210?1?1?e2?0tee21?t22dt???0e1?t22td(?)??e22?
(16)?1 解
dxx1?lnxdxx1?lnxdx2?
e21?1e2??1e211?lnxdlnx?21?lnx?2(3?1)?
(17)?0x?2x?2
00dx1 解 ??22???2dx?arctan(x?1)2x?2x?21?(x?1) (18)?2?cosxcos2xdx?
?2?2?
0?2?arctan1?arctan(?1)??2?
?
解
?2cosxcos2xdx?2(1?2sin2????22???2x)dsinx?(sinx?sin3x)3?2??22?? 3
(19)?2?cosx?cos3xdx?
?2?
解
??2??2cosx?cosxdx??2?cosx1?cos2xdx?23?
????cosx(?sinx)dx??20?2032cosxsinxdx?cos2x30??232?cos2x3?204? 3 (20)?01?cos2xdx? 解
?
?0?0??1?cos2xdx?2?0sinxdx??2cosx??22?
2? 利用函数的奇偶性计算下列积分? (1)???x4sinxdx?
? 解 因为x 4sin x在区间[??? ?]上是奇函数? 所以???x4sinxdx?0? (2)?2?4cos4?d??
?2
?
解
?24cos4?d???2??2?24cos4?d?0??8?02(??1?cos2x2)d?2
?2?2(1?2cos2x?cos20?312x)d??2?02(?2cos2x?cos4x)d?
22?201 ?(3??2sin2x?sin4x)4?3?? 2
(3)?12?12(arcsinx)21?x2dx?
解
?12?12(arcsinx)21?x2dx?2?120(arcsinx)21?x2dx?2?02(arcsinx)2d(arcsinx)1
12?332 ?(arcsinx)? ?03324325xsinxdx? (4)??54 2x?2x?1325xsinxx3sin2xdx?0? 解 因为函数4是奇函数? 所以??5422x?2x?1x?2x?1
3? 证明?
22??a?(x)dx?2?0?(x)dx? 其中?(u)为连续函数?
aa
证明 因为被积函数?(x2)是x的偶函数? 且积分区间[?a? a]关于原点对称? 所以有
22??a?(x)dx?2?0?(x)dx? aa
4? 设f(x)在[?b? b]上连续? 证明??bf(x)dx???bf(?x)dx?
bb 证明 令x??t? 则dx??dt? 当x??b时t?b? 当x?b时t??b? 于是 而 所以
??bf(x)dx??bbb?bf(?t)(?1)dt???bf(?t)dt?
b
??bf(?t)dt???bf(?x)dx? ??bf(x)dx???bf(?x)dx?
bbb
5? 设f(x)在[a? b]上连续?? 证明?af(x)dx??af(a?b?x)dx?
bb 证明 令x?a?b?t? 则dx?d t? 当x?a时t?b? 当x?b时t?a? 于是 而 所以
?af(x)dx??bf(a?b?t)(?1)dt??af(a?b?t)dt? ?af(a?b?t)dt??af(a?b?x)dx? ?af(x)dx??af(a?b?x)dx?
bbbbbab
6? 证明?
(x?0)?
1?x2111 证明 令x?? 则dx??2dt? 当x?x时t?? 当x?1时t?1? 于是
txt?x1?x21dx??1x1dx
1111dx1x??(?)dt?dt? ?x?2?122111?xt1?tx1?2t1
而
1?x21dxx所以 ?x? ?2?211?x1?x
?1x111?t21dxdt??1x11dx?
7? 证明?
?1m1nnmx(1?x)dx?x(1?x)dx? 00?
1 证明 令1?x?t ? 则?0xm(1?x)ndx???1(1?t)mtndt??0(1?t)mtndt??0xn(1?x)mdx?
101
同济大学第六版高等数学上下册课后习题答案5-3
