计算不定积分∫xsinxcos5xdx

计算不定积分∫xsinxcos5xdx.
2025-05-14 23:44:07
推荐回答(1个)
回答1:

因为

sinx
cos5x
=(
1
4cos4x
)′
,所以
xsinx
cos5x
dx

=
1
4
∫xd(
1
cos4x
)

=
1
4
(
x
cos4x
?∫
dx
cos4x
)

又因为
dx
cos4x
 
=
sin2x+cos2x
cos2x
?
dx
cos2x

=∫(tan2x+1)d(tanx)
=
1
3
tan3x+tanx
-C,
所以
xsinx
cos5x
dx
=
1
4
(
x
cos4x
?
1
3
tan3x?tanx+C)
=
x
4cos4x
?
1
12
tan3x?
1
4
tanx+C