8.2.5.evaluate int 64x sec^2(4x) dx

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Discussion Overview

The discussion revolves around the evaluation of the integral \( I = \int 64x \sec^2(4x) \, dx \). Participants explore methods of integration, particularly integration by parts, and discuss the implications of different approaches and results.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant begins by stating the integral to evaluate and expresses uncertainty about which trigonometric identity to use.
  • Another participant suggests using integration by parts, providing specific substitutions for \( u \) and \( dv \), and derives part of the integral's solution.
  • A similar response reiterates the integration by parts approach and provides a detailed calculation leading to an expression involving the logarithm of the cosine function.
  • Participants express uncertainty about the correctness of the final result, particularly in relation to the absolute value in the logarithmic term.
  • One participant questions whether the answer provided by Wolfram Alpha differs from their own, indicating a potential discrepancy in results.
  • Another participant notes that Wolfram Alpha omitted the absolute value around the logarithmic argument, raising questions about the reasoning behind this omission.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final result of the integral. There are competing views regarding the inclusion of absolute values in the logarithmic expression, and uncertainty remains about the correctness of the derived solution.

Contextual Notes

Participants do not clarify certain assumptions or dependencies in their calculations, particularly regarding the treatment of the logarithmic function and its arguments. The discussion reflects a reliance on integration techniques without resolving all mathematical steps.

karush
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Evaluate $I=\displaystyle \int 64x\sec^2(4x) \, dx$

ok well first $64 \displaystyle\int x\sec^2(4x) \, dx$

off hand not sure what trig id to use or if we need to
 
Last edited:
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integration by parts …

$u = 16x \implies du = 16 \, dx$
$dv = 4\sec^2(4x) \, dx \implies v = \tan(4x)$

$\displaystyle \int 16x \cdot 4\sec^2(4x) \, dx = 16x \cdot \tan(4x) - \int 16\tan(4x) \, dx$

can you finish up?
 
skeeter said:
integration by parts …

$u = 16x \implies du = 16 \, dx$
$dv = 4\sec^2(4x) \, dx \implies v = \tan(4x)$

$\displaystyle \int 16x \cdot 4\sec^2(4x) \, dx = 16x \cdot \tan(4x) - \int 16\tan(4x) \, dx$

can you finish up?
so if ... (from plug in table)
$\displaystyle 16 \int \tan \left(4x\right)dx
=16\left[-\dfrac{1}{4}\ln \left|\cos \left(4x\right)\right|\right]+C
=-4\ln \left|\cos \left(4x\right)\right|+C$
Hence
$I=16x \cdot \tan(4x)+4\ln \left|\cos \left(4x\right)\right|+C$
Screenshot 2021-12-20 8.39.08 PM.png

anyway not sure...
 
not sure about what?
 
isn't W|A answer different or is it??
 
W/A omitted the absolute value around the log argument.
Why it did, I don't know.
 

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