Integration
posted by Vidal .
Intergrate ¡ì sec^3(x) dx
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^{3}(x) into a piece of software and got an answer that looks slightly different from yours, but I'm not positive. If you still need help with this post a new question so it's easy to find.

s=integral
let i=S(sec^3x)dx
i=S(sec^3x)=S(sec^2x.secx).dx
=tanxsecxS(tan^2x.secx)dx=
tanxsecxS(sec^2x1)secx.dx=
tanxsecxS(sec^3x)dx+S(secx)dx
2i=tanxsecx+lnsecx+tanx+x
i=(tanxsecx+lnsecx+tanx+c)(0.5) 
cos(2x)upon2