求secθ대的不定积分(请写过程)

2024-11-25 04:55:36
推荐回答(1个)
回答1:


sec³xdx
=∫
secxd(tanx)
=secxtanx-∫
tan²xsecxdx
=secxtanx-∫
(sec²x-1)secxdx
=secxtanx-∫
sec³xdx+∫secxdx
=secxtanx-∫
sec³xdx+ln|secx+tanx|
将-∫
sec³xdx移动等式左边与左边合并,除去系数(别忘记要留常数C在右边)

sec³xdx=(1/2)secxtanx+(1/2)ln|secx+tanx|+C
参考:

sec³x
dx
=

secx*sec²x
dx
=

secx
dtanx,分部积分法,sec²x的积分是tanx
=
secx*tanx
-

tanx
dsecx,分部积分法
=
secx*tanx
-

tanx*(secx*tanx)
dx,secx的导数是secx*tanx
=
secx*tanx
-

secx(sec²x-1)
dx,恒等式1+tan²x
=
sec²x
=
secx*tanx
-

sec³x
dx
+

secx
dx,将∫sec³x
dx移到等号左边,变为2个∫
sec³x
dx
2∫sec³x
dx
=
secx*tanx
+

secx*(secx+tanx)/(secx+tanx)
dx,上下分别乘以secx+tanx

sec³x
dx
=
(1/2)secx*tanx
+
(1/2)∫
(secx*tanx+sec²x)/(secx+tanx)
dx
=
(1/2)secx*tanx
+
(1/2)∫
d(secx+tanx)/(secx+tanx),等同公式∫
1/u
du,u=secx+tanx
∴∫
sec³x
dx
=
(1/2)secx*tanx
+
(1/2)ln|secx+tanx|
+
C
递推公式:

(secx)^n
dx
=
[sinx*(secx)^(n-1)]/(n-1)
+
(n-2)/(n-1)*∫
(secx)^(n-2)
dx