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Integral of sec(x)^3 
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#1
Feb1307, 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 


#2
Feb1307, 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 + Lnsecx + tanx INT[sec^3x]dx = 1/2 secx.tanx + 1/2 Lnsecx + tanx 


#3
Feb1407, 12:43 AM

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P: 11,948

[tex]\int \frac{\cos x}{\left( 1\sin ^{2}x\right) ^{2}}dx=\allowbreak \int \frac{1}{\left( 1t^{2}\right) ^{2}}\,dt [/tex]
for the last integral, use simple fractions. [itex] \sin x=t [/itex] has been used. 


#4
Feb1407, 07:26 AM

P: 54

Integral of sec(x)^3
Ah, alright thanks. It was simpler than i thought it would be



#5
Feb1407, 07:55 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

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}{(1u^2)^2}= \int\frac{du}{(1u)^2(1+u)^2}[/tex] and, again, use partial fractions. 


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