Integrating the product of a step function and a trigonometric function

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SUMMARY

The integral of the product of a step function and a trigonometric function, specifically -π∫π H(t-π/2)*sin(2t)dt, evaluates to 0 due to the properties of the unit step function H(t). The function H(t-π/2) equals 0 for t < π/2 and 1 for t ≥ π/2, thus simplifying the integral to ∫(π/2 to π) sin(2t) dt. This confirms that the integral over the specified limits yields a non-zero value, contradicting the initial assumption that it would be 0.

PREREQUISITES
  • Understanding of unit step functions, specifically H(t).
  • Knowledge of trigonometric functions, particularly sin(2t).
  • Familiarity with definite integrals and their properties.
  • Basic calculus concepts, including integration techniques.
NEXT STEPS
  • Study the properties of the unit step function H(t) in detail.
  • Learn how to evaluate definite integrals involving trigonometric functions.
  • Explore integration techniques for piecewise functions.
  • Investigate the implications of shifting functions in integration.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques involving step functions and trigonometric functions.

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Homework Statement



Integrate π H(t-π/2)*sin(2t)dt

Homework Equations



See above.

The Attempt at a Solution



I can rationalize the slightly simpler integral for the same limits of H(t)*sin(2t) as coming out to 0 due to the definition of the unit step function, but I'm wondering if the subtraction of π/2 changes it any. It's still integrating over the range of H(t), correct? So should it still work out to be 0?
 
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[itex]H(t- \pi/2)[/itex] is equal to 0 for [itex]t< \pi/2[/itex], 1 for [itex]t\ge \pi/2[/itex]. So that integral is just
[tex]\int_{\pi/2}^\pi sin(2t)dt[/tex]
 
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