The integral of (secx)^3 tan x dx.
2006-12-02 04:52:24 · 4 answers · asked by ANON 1 in Science & Mathematics ➔ Mathematics
sec^xtanxdx
=sec^x*secxtanxdx
put secx=t
secxtanxdx=dt
the integral=t^2dt
=t^/3+C
=sec^x/3+C
2006-12-02 05:00:09 · answer #1 · answered by raj 7 · 0⤊ 0⤋
Write the integrand as (sec²x )sec x tan x dx.
Now what do you notice? You have the derivative
of sec x tan x here. So let u = sec x; du = sec x tan x dx
We get int[u² du] = u^3/3 + C = sec^3 x/3 + C.
2006-12-02 14:42:04 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
(secx)^3 tanx dx
u = secx
du = secxtanx dx
(secx)^3 tanx dx = (secx)^2(secx)(tanx) dx
= u^2du
Integrate: u^3/3
= (secx)^3/3 + C
2006-12-02 12:57:42 · answer #3 · answered by Morkeleb 3 · 0⤊ 1⤋
Do your own homework!
2006-12-02 12:56:43 · answer #4 · answered by campmor 3 · 0⤊ 2⤋