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January 29, 2015

January 29, 2015

Posted by **Vidal** on Sunday, October 15, 2006 at 2:06am.

could anybody please check this answer. are the steps correct? thanks.

= ¡ì sec x d tan x

= sec x tan x - ¡ì tan x d sec x

= sec x tan x - ¡ì sec x tan^2(x) dx

= sec x tan x + ¡ì sec x dx - ¡ì sec^3(x) dx

= sec x tan x + ln |sec x + tan x| - ¡ì sec^3(x) dx

=¡ì sec^3(x) dx = (1/2)(sec x tan x + ln |sec x + tan x|) + C1

¡ì [3x sin x/cos^4(x)] dx

= -3 ¡ì [x/cos^4(x)] d cos x

= ¡ì x d sec^3(x)

= x sec^3(x) - ¡ì sec^3(x) dx

= x sec^3(x) - (1/2) sec x tan x - (1/2) ln |sec x + tan x| + C2

I'm not sure if your integration is correct or not, not all of your symbols converted to ASCII. I plugged sec

- egypt-german univerty in cairo-sherif.el-sayed@student.guc.edu.eg -
**sherif**, Tuesday, March 4, 2008 at 12:07pms=integral

let i=S(sec^3x)dx

i=S(sec^3x)=S(sec^2x.secx).dx

=tanxsecx-S(tan^2x.secx)dx=

tanxsecx-S(sec^2x-1)secx.dx=

tanxsecx-S(sec^3x)dx+S(secx)dx

2i=tanxsecx+ln|secx+tanx|+x

i=(tanxsecx+ln|secx+tanx|+c)(0.5)

- lstem jxbsi -
**lstem jxbsi**, Sunday, February 1, 2009 at 3:19pmwpjceq gcifns hoxnis ytxuwoz plfm zuonie kwhbemp

- lstem jxbsi -

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