Solving Integral (sec^3 X) Quickly

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SUMMARY

The integral of secant cubed, represented as I = ∫ sec³x dx, can be solved using integration by parts. By letting u = sec x and dv = sec²x dx, the solution simplifies to I = sec x tan x - ∫ sec x tan²x dx. Utilizing the identity tan²x = sec²x - 1, the integral is further reduced, leading to the final result: I = (1/2)[sec x tan x + ln|tan x + sec x|] + C. Additional resources for derivatives and integrals include Calc101 and Compute UW-Lax.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with trigonometric identities, particularly tan²x = sec²x - 1.
  • Basic knowledge of calculus, including antiderivatives and their applications.
  • Experience with mathematical notation and functions in programming contexts.
NEXT STEPS
  • Study the method of integration by parts in detail.
  • Learn more about trigonometric identities and their applications in calculus.
  • Explore advanced integration techniques, such as substitution and partial fractions.
  • Investigate online tools for calculus, including Wolfram Integrals and Calc101.
USEFUL FOR

Students and educators in calculus, mathematicians seeking to refine integration skills, and anyone looking for efficient methods to solve trigonometric integrals.

Gale
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Hey, i know this is an easy problem, but i don't have any of my older calc books with the identities, so, just hoping for a quick answer

integral(sec^3 X)

thanks... i just need that for a more difficult problem... so i don't want to spend a lot on this simple integral, thanks.
 
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let I = \int sec^3xdx

and let

u = secx \ \ \Rightarrow \ \ du = secxtanxdx

dv = sec^2x \ \ \Rightarrow \ \ v = tanx

using integration by parts we have

I = uv - \int vdu

\Rightarrow \ \ I = secxtanx - \int secxtan^2xdx

using the trigonometric identity tan^2 = sec^2 -1 the integral on the right side becomes
I = secxtanx - \int[secx(sec^2x - 1)]dx

\Rightarrow I = secxtanx - \int sec^3xdx + \int secxdx

\Rightarrow I = secxtanx - I + ln|tanx + secx|

\Rightarrow 2I = secxtanx + ln|tanx + secx|

\Rightarrow I = \frac{1}{2}[ \ secxtanx + ln|tanx + secx| \ ] + C
 
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Derivatives are easy for all products, quotients, powers, compositions, and sums of elementary functions... antiderivatives are much more difficult.
 
tony873004 said:
That's a great link :cool:

Is there one that does derivatives?
YES ... Integrals, Derivatives, Graphing, and Matrices ... GO TO:
-----> http://www.calc101.com/
Select from list upper right corner home page. Remember to enter functions with SQUARE BRACKETS "[]", like "sin[x]", and group with PARENTHESES "()", like:
"(x + 3*exp[5*x])*(x*sin[x])^3"



~~
 
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