what is the integral of...?

Question

what is the integral of (secx)^3? secant cubed x spelled out.

Answer 1

integral by parts with
u=secx dv=sec^2xdx

du=secxtanxdx v=tanx

so int(sec^3xdx)=left hand side
i just call it. you will see a skill n this one
LHS= secxtanx-int(secxtan^2x)dx
[use part like uv-int(vdu)]
LHs= secxtanx-int(secx(sec^2x-1)dx)
if you multiply the second integral we have
LHS=secxtanx-int sec^3xdx+int(secx)dx
that is skill providing
add both sides by int(sec^3xdx)
LHS+int(sec^3xdx)= secxtanx+int(secxdx)
but LHs=intsec^3xdx
put it in
2intsec^3xdx= secxtanx+inf(secxdx)
divide 2 both sides
int(sec^3xdx)= 1/2(secxtanx+int(secx)dx)
int(sec^3xdx)= 1/2(secxtanx+ln|secx+tanx|)+C
that is answer
[[[[[[[[[if you want me to show you how to get int secxdx
( int(secxdx). time top and bottom secx+tanx
so int secx(secx+tanx)dx/(secx+tanx)
let u=secx+tanx
du=(secxtanx+sec^2x)dx
if on the top you multiply you will get secxtanx+sec^2x
so int du/u= ln|u|+c
so but u= secx+tanx back
ln|secx+tanx| +C]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]
answer is
int(sec^3xdx)= 1/2(secxtanx+ln|secx+tanx|)+C
good luck.

Answer 2

let u = sec x , du = sec x tan x dx
dv = (sec x)^2 , v = tan x

integral of (secx)^3 dx= secxtanx - integral of secx(tanx)^2 dx
=secxtanx - integral of secx( (secx)^2 - 1) dx
=secxtanx - integral of (secx)^3 dx +
integral of (secx)dx

Using integal of sec x = ln(secx + tanx) + c, we get

integral of secx^3 = 1/2 * ( secxtanx + ln(secx + tanx) ) + c

Answer 3

The integral of (sec(x))^3 is:

(1/4)ln(|sin(x)+1|) - (1/4)ln(|sin(x)-1|) - ((1/4)/(sinx-1)) - ((1/4)(sinx+1))

Perhaps you can factor out some terms there... but I leave that part to you :)