Simple closed form for integral

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SUMMARY

The integral $$\int_{0}^{1}t\cos(2t\pi)\tan(t\pi)\ln[\sin(t\pi)]\mathrm dt$$ evaluates to $$\frac{1}{\pi}\cdot\frac{\ln 2}{2}(1-\ln 2)$$. This conclusion is reached through specific techniques in calculus involving trigonometric identities and logarithmic properties. The discussion emphasizes the need for a structured approach to solving complex integrals, particularly those involving oscillatory functions and logarithms.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with trigonometric functions and identities
  • Knowledge of logarithmic properties
  • Experience with definite integrals
NEXT STEPS
  • Study techniques for evaluating integrals involving trigonometric functions
  • Learn about the properties of logarithmic integrals
  • Explore advanced calculus topics such as Fourier series
  • Investigate numerical methods for approximating complex integrals
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Mathematicians, calculus students, and anyone interested in advanced integral evaluation techniques.

Tony1
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How may we go about to show that,

$$\int_{0}^{1}t\cos(2t\pi)\tan(t\pi)\ln[\sin(t\pi)]\mathrm dt=\color{green}{1\over \pi}\cdot\color{blue}{{\ln 2\over 2}(1-\ln 2)}$$
 
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A hint is requested ... (Blush)
 
lfdahl said:
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