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Integral of sec(x)^3

  • Thread starter imranq
  • Start date
  • #1
57
1

Homework Statement



I'm just trying to solve this

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

Homework Equations


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


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
 
Last edited:

Answers and Replies

  • #2
30
0
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|
 
  • #3
dextercioby
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[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.
 
  • #4
57
1
Ah, alright thanks. It was simpler than i thought it would be
 
  • #5
HallsofIvy
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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.
 

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