Register to reply

Integral of sec(x)^3

by imranq
Tags: integral, secx3
Share this thread:
imranq
#1
Feb13-07, 05:52 PM
P: 54
1. The problem statement, all variables and given/known data

I'm just trying to solve this

[tex] \int {\sec^3{x} dx [/tex]

2. Relevant equations
[tex] sec^2{x} = 1 + tan^2{x} [/tex]


3. The attempt at a solution

well i was able to simplify it to this:

[tex] \int {\sec{x}*\tan^2{x}} dx + \ln{|\sec{x} + \tan{x}|} [/tex]

but I still was not able to find that new integral
Phys.Org News Partner Science news on Phys.org
New model helps explain how provisions promote or reduce wildlife disease
Stress can make hard-working mongooses less likely to help in the future
Grammatical habits in written English reveal linguistic features of non-native speakers' languages
SunGod87
#2
Feb13-07, 08:14 PM
P: 30
INT[sec^3x]dx
= INT[secx.sec^2x]dx
By parts (differentiating secx and integrating sec^2x):
= INT[secx.sec^2x]dx
= secx.tanx - INT[secx.tan^2x]dx
= secx.tanx - INT[secx(sec^2x - 1)]dx
= secx.tanx - INT[sec^3x - secx]dx
So we have:

INT[sec^3x]dx = secx.tanx - INT[sec^3x - secx]dx
INT[sec^3x]dx = secx.tanx - INT[sec^3x]dx + INT[secx]dx
2 INT[sec^3x]dx = secx.tanx + INT[secx]dx
2 INT[sec^3x]dx = secx.tanx + Ln|secx + tanx|
INT[sec^3x]dx = 1/2 secx.tanx + 1/2 Ln|secx + tanx|
dextercioby
#3
Feb14-07, 12:43 AM
Sci Advisor
HW Helper
P: 11,894
[tex]\int \frac{\cos x}{\left( 1-\sin ^{2}x\right) ^{2}}dx=\allowbreak \int \frac{1}{\left( 1-t^{2}\right) ^{2}}\,dt [/tex]

for the last integral, use simple fractions. [itex] \sin x=t [/itex] has been used.

imranq
#4
Feb14-07, 07:26 AM
P: 54
Integral of sec(x)^3

Ah, alright thanks. It was simpler than i thought it would be
HallsofIvy
#5
Feb14-07, 07:55 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,301
Another way: Since sec(x)= 1/cos(x),
[tex]\int sec^3(x)dx= \int\frac{dx}{cos^3(x)}[/tex]
which is an odd power of cos(x). Multiply numerator and denominator by cos(x):
[tex]\int \frac{cos(x)dx}{cos^4(x)}= \int\frac{cos(x)dx}{(1- sin^2(x))^2}[/tex]
Let u= sin(x) so du= cos(x)dx
[tex]\int \frac{du}{(1-u^2)^2}= \int\frac{du}{(1-u)^2(1+u)^2}[/tex]
and, again, use partial fractions.


Register to reply

Related Discussions
Integral more general then Lebesgue integral? Calculus 7
Integral - Double integral Calculus & Beyond Homework 2
Need help w/ an integral Calculus 4
QM Integral and Online Integral Tables Calculus 5
Integral using integral tables Calculus 2