# Long Integral equals (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π)

• MHB
• Tony1
In summary: Using the power rule and the product rule for integration again, we can expand the integral to:$$\frac{1}{\pi}\int_{0}^{\pi}\left(u^2\sec(u)+u^2\sec(u)\sin(u)-\frac{1}{2}u^2\sec(u)\right)\ln[\sin(u)]\mathrm du$$Now, we can use the substitution $v = \sin(u)$ and $dv = \cos(u)\mathrm du$ to further simplify the integral:$$\frac{1}{\pi}\int_{0}^{1}\ Tony1 How to show that,$$\int_{0}^{1}t^2\sec(t\pi)[1+\sin(t\pi)-\sin^2(t\pi)]\ln[\sin(t\pi)]\mathrm dt=\left({\pi\over 2}\right)^2-{\color{blue}{2\ln(2)\over \pi^2}}+\color{red}{{\ln^2(2)\over 2\pi}}$$Hello, To show this integral, we can start by using the trigonometric identity \sin^2(t\pi) = \frac{1}{2}(1-\cos(2t\pi)). Then, we can rewrite the integral as:$$\int_{0}^{1}t^2\sec(t\pi)\left[1+\sin(t\pi)-\frac{1}{2}(1-\cos(2t\pi))\right]\ln[\sin(t\pi)]\mathrm dt$$Next, we can use the substitution u = t\pi and du = \pi\mathrm dt to simplify the integral:$$\frac{1}{\pi}\int_{0}^{\pi}u^2\sec(u)\left[1+\sin(u)-\frac{1}{2}(1-\cos(2u))\right]\ln[\sin(u)]\mathrm du$$Using the power rule and the product rule for integration, we can expand the integral to:$$\frac{1}{\pi}\int_{0}^{\pi}\left(u^2+u^2\sin(u)-\frac{1}{2}(u^2-u^2\cos(2u))\right)\sec(u)\ln[\sin(u)]\mathrm du$$Simplifying further, we get:$$\frac{1}{\pi}\int_{0}^{\pi}\left(u^2+u^2\sin(u)-\frac{1}{2}u^2+\frac{1}{2}u^2\cos(2u)\right)\sec(u)\ln[\sin(u)]\mathrm du$$Using the trigonometric identity \cos(2u) = 1-2\sin^2(u), we can rewrite the integral as:$$\frac{1}{\pi}\int_{0}^{\pi}\left(\frac{1}{2}u^2+u^2\sin(u)-\frac{1}{2}u^2+u^2\sin^2(u)\right)\sec(u)\ln[\sin(u)]\mathrm du$$Simplifying further, we get:$$\frac{1}{\pi}\int_{0}^{\pi}\left(u^2+u^2\sin(u)-\frac{

## 1. What is the meaning of "Long Integral equals (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π)"?

The expression "Long Integral equals (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π)" is a mathematical formula that represents a long integral calculation. The result of this calculation is a numerical value.

## 2. How do you solve the long integral in (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π)?

The long integral in (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π) can be solved using various methods such as integration by parts, substitution, or partial fractions. The specific method used may depend on the complexity of the integral and the available tools at hand.

## 3. Can the result of the long integral in (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π) be simplified further?

Yes, the result of the long integral can be simplified further using algebraic manipulation or other mathematical techniques. However, the level of simplification may vary depending on the specific integral and the desired level of precision.

## 4. What is the significance of the number (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π) in mathematics?

The number (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π) does not have any specific significance in mathematics. It is simply the result of a long integral calculation and can be used in further calculations or as a numerical value in a mathematical context.

## 5. Is there any real-world application of the long integral in (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π)?

The long integral in (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π) may have applications in various fields such as physics, engineering, and economics. It can be used to calculate various quantities and solve problems that involve continuous functions.

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