Where are you getting these monstrosities from?Tony said:How to prove this integral,
$$\int_{0}^{2\pi}\sin\left({x\over 2}\right)\ln^2\left[\sin\left({x\over 4}\right)\sin\left({x\over 8}\right)\right]={27\over 2}+\ln^2(2)+\ln(2)$$
greg1313 said:Hi Tony and welcome to MHB! :D
Why are you posting these problems? Also, do you have the solutions?
I am considering moving them to the "Challenge Questions an Puzzles" forum and I can mark them as "Unsolved Challenges" if you do not have solutions.
Also, I encourage you to give more meaningful titles to your threads - I will be renaming several of them in the near future.
Good evening,
greg1313
The integral of sine is the function that represents the area under the curve of the sine function from 0 to a given point on the x-axis. It is represented by the formula ∫sin(x) dx.
To solve for the integral of sine, you can use integration techniques such as u-substitution, integration by parts, or trigonometric identities. In this particular case, the integral of sine equals 27/2+ln^2(2)+ln(2).
The constant 27/2+ln^2(2)+ln(2) represents the value of the definite integral of sine from 0 to a given point on the x-axis. It is the numerical value of the area under the curve of the sine function.
The values in the natural logarithm represent the ratio between the area under the curve of the sine function and the width of the interval from 0 to a given point on the x-axis. In other words, it represents the natural logarithm of the base of the exponential function that would give the same result as the integral of sine.
Yes, the integral of sine can be negative if the area under the curve of the sine function is below the x-axis. In this case, the value of the integral would be the negative of the numerical value of the area. However, in the given equation, the integral of sine equals a positive value, 27/2+ln^2(2)+ln(2), which represents the area above the x-axis.