Intergrate ¡ì sec^3(x) dx

Question

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 sec³(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.

Answer 1

s=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)

Answer 2

-cos(2x)upon2