Integtrating secant and tangent

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Homework Help Overview

The discussion centers around the integration of the function sec^3(2x) * tan(2x) with respect to x. Participants are exploring various approaches to tackle this integral within the context of calculus.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • One participant attempts to separate the secant and tangent functions and uses substitution, expressing concerns about the correctness of their approach. Another participant suggests a different substitution and questions the derivative of a new variable introduced. A third participant proposes an alternative method by splitting the integrand into components, indicating a preference for a simpler approach.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the validity of their methods. There is no explicit consensus, but multiple strategies are being explored, indicating a productive exchange of ideas.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the amount of direct assistance they can provide to one another. There are indications of uncertainty regarding the correct application of integration techniques and substitutions.

Aerosion
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Homework Statement



(integrate)sec^3(2x)*tan(2x)dx

Homework Equations





The Attempt at a Solution



Okay, so first I tried to separate secant and tangent, to get (integrate)1/cos^3(2x) * sin(2x)/cos(2x)dx. The u would be cos(2x), the du would be -du=sin(2x)dx.

I subsitute these in the aforementioned equation, whcih leaves (integrate)1/u^3 x 1/u. The problem is, I don't think this is right.
 
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u = 2x
du = 2 dx

[itex]\frac 12 \int \sec ^3 u \tan u du[/itex]

Now let [itex]v = \sec ^2 u[/itex] What is dv?
 
Aerosion said:

Homework Statement



(integrate)sec^3(2x)*tan(2x)dx

Homework Equations





The Attempt at a Solution



Okay, so first I tried to separate secant and tangent, to get (integrate)1/cos^3(2x) * sin(2x)/cos(2x)dx. The u would be cos(2x), the du would be -du=sin(2x)dx.

I subsitute these in the aforementioned equation, whcih leaves (integrate)1/u^3 x 1/u. The problem is, I don't think this is right.

There's the mistake.
 
Slight change to IMDereks post. I don't know wat dv is, can't be bothered working it out, i think my methods quicker. let v= sec u, split the integrand into sec^2 u (sec u tan u), that way the things in the brackets it dv/du, the integral is now integral: v^2 dv. simpler i think
 

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