}^{\ln(4)}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}dx&line2=\mathrm{Solution:\:}\arcsin(\frac{\sqrt{15}}{4})-\frac{%CF%80}{3} "What is the integral from ln(2) to ln(4) of (e^{-x})/(sqrt(1-e^{-2x))} ?")
What is the integral from ln(2) to ln(4) of (e^{-x})/(sqrt(1-e^{-2x))} ?
Pi Approximations -- from Wolfram MathWorld
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e^pi - pi + 9/10^4 + 1/(10^4*ln(2) + sqrt(10)/6)^2
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int -pi/4^pi/4ln√(1 + sin2x)dx has the value equal to:
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SOLVED: Use a double integral to find the area of the region within the circle x 2 + y 2 = 2 and above the line y = 1.
Improper Integrals
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Question Corner -- Why is e^(pi*i) = -1?
