Anti-derivative of a trig equation

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SUMMARY

The discussion focuses on finding the anti-derivative of the equation \(\int x(\sin^2(3x)\cos(3x))dx\). The recommended approach involves using integration by parts, specifically setting \(u = x\) and \(dv = \sin^2(3x)\cos(3x)dx\). Additionally, a substitution \(y = \sin(3x)\) is suggested to simplify the integration of \(\sin^2(3x)\cos(3x)dx\). This method effectively addresses the problem posed by the original poster.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with trigonometric identities, particularly \(\sin^2(x) + \cos^2(x) = 1\).
  • Knowledge of substitution methods in calculus.
  • Basic proficiency in handling integrals involving trigonometric functions.
NEXT STEPS
  • Study the integration by parts technique in detail, focusing on its applications.
  • Learn how to apply trigonometric identities to simplify integrals.
  • Explore substitution methods in calculus, particularly for trigonometric functions.
  • Practice solving integrals involving products of polynomial and trigonometric functions.
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Students studying calculus, particularly those tackling integration problems involving trigonometric functions, as well as educators looking for effective teaching strategies in integration techniques.

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Homework Statement


I just need to find the anti-derivative of this equation:
\int x(sin^2(3x)cos(3x))dx



Homework Equations


sin^2(x)+cos^2(x)=1


The Attempt at a Solution


I'm not really sure where to start here. I tried to do a substitution first but couldn't make it work. So then I tried to put sin^2 in terms of cos^2 but that didn't seem to help me out any more. Anyone have an idea of how to tackle this problem? Thanks in advance!
 
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togame said:

Homework Statement


I just need to find the anti-derivative of this equation:
\int x(sin^2(3x)cos(3x))dx



Homework Equations


sin^2(x)+cos^2(x)=1


The Attempt at a Solution


I'm not really sure where to start here. I tried to do a substitution first but couldn't make it work. So then I tried to put sin^2 in terms of cos^2 but that didn't seem to help me out any more. Anyone have an idea of how to tackle this problem? Thanks in advance!

Integrate by parts, with u = x, dv = \sin^2(3x)\cos(3x)dx.

To integrate \sin^2(3x)\cos(3x)dx, use the substitution y = \sin(3x)
 
Curious3141 said:
Integrate by parts, with u = x, dv = \sin^2(3x)\cos(3x)dx.

To integrate \sin^2(3x)\cos(3x)dx, use the substitution y = \sin(3x)

Thank you very much. I didn't see the ability to do another substitution right off hand.
 

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