Integral: š›¼^(2x)cos(x)dx Solution

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SUMMARY

The integral of š›¼^(2x)cos(x)dx can be solved using integration by parts, specifically by applying the technique twice. This method leads to a term that resembles the original integral, allowing for the collection of terms to isolate and solve for the integral. The discussion highlights the importance of persistence in problem-solving and the effectiveness of integration by parts in tackling complex integrals.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts
  • Familiarity with u-substitution in calculus
  • Knowledge of basic integral properties and functions
  • Experience with calculus textbooks, such as Larson's Calculus
NEXT STEPS
  • Practice solving integrals using integration by parts with various functions
  • Explore advanced integration techniques covered later in calculus textbooks
  • Review examples of integrals that require multiple applications of integration by parts
  • Study the concept of collecting terms in integrals to simplify complex expressions
USEFUL FOR

Students studying calculus, particularly those struggling with integration techniques, and educators seeking effective methods to teach integration by parts.

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Homework Statement


\inte^{2x}cos(x)dx

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The Attempt at a Solution


This isn't exactly a homework question, but it's close enough: it's from my old Calculus book (Larson), and I've been stuck on it for awhile. I just need a little hint or something. I've tried a u-substitution and haven't found anything that seems to help, and I've tried integration by parts, but nothing really simplifies--I just end up with an integral that is just as hard as before I try integration by parts. Maybe I'm using these techniques wrong or not seeing something obvious. No other 'harder' integration techniques should be needed (they come later in the book). Thanks!
 
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Try integrating by parts twice. You'll eventually end up with a term that looks like the original integral. You can collect terms, so to speak, and solve for the integral.
 
Ah! I would've never thought of that. Thanks!
 

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