Evaluating the Definite Integral of ln2(cos(x)) from 0 to π/2

Integral calculus plays a crucial role in solving complex integrals, especially those involving trigonometric functions. One such intriguing problem is the evaluation of the integral of ln2(cos(x)) from 0 to π/2. This integral can be solved using a combination of trigonometric identities and techniques such as symmetry properties and substitution.

Using Symmetry Properties and Trigonometric Identities

Consider the integral:
I ∫?^(π/2) ln2(cos(x)) dx

To evaluate this integral, we start by using a symmetry property of definite integrals. Notice that cos(π/2 - x) sin(x). This property allows us to write:

_integral from 0 to π/2 f(x) dx integral from 0 to π/2 f(π/2 - x) dx_

Applying this to our integral:

I ∫?^(π/2) ln2(cos(x)) dx ∫?^(π/2) ln2(sin(x)) dx

By adding these two equal integrals, we obtain:

2I ∫?^(π/2) ln2(cos(x)) dx ∫?^(π/2) ln2(sin(x)) dx

Recognizing that cos2(x) * sin2(x) 1/4 * sin2(2x), we can proceed to evaluate the integral. Let u ln cos(x), then du -2 sin(x) cos(x) dx. The integral transforms to:

∫?? ln(u^0.5)2 - du/ -1 - u^0.5 ∫?1 ln(u)2 / (1 - u) du

This integral is well-known and can be evaluated using standard methods or from a table of integrals, resulting in:

∫?1 ln(u)2 / (1 - u) du -4π3/27

Therefore, we have:

I 1/2 ∫?1 ln(u)2 / (1 - u) du 1/2 * (-4π3/27) -2π3/27

Using Fourier Series

A more sophisticated approach involves the use of the Fourier series for ln(cos(x)). The Fourier series for ln(cos(x)) is given by:

ln(cos(x)) -ln(2) Σ???^(∞) (-1)^(n-1) / n * cos(2nx)

Squaring this series, we obtain:

ln2(cos(x)) ln2(2) - 2 ln(2) * Σ???^(∞) (-1)^(n-1) / n * cos(2nx) * Σ???^(∞) (-1)^(m-1) / m * cos(2mx)

Integrating both sides from 0 to π/2 and using the results:

∫?^(π/2) cos(2nx) dx 0

∫?^(π/2) cos(2mx) cos(2nx) dx 0 if m ≠ n and π/4 if m n

We find:

∫?^(π/2) ln2(cos(x)) dx ln2(2) * π/2 - 2 ln(2) * 0 π/4 * Σ???^(∞) 1/n2

The last sum is a known series, Σ???^(∞) 1/n2 π2/6, resulting in:

∫?^(π/2) ln2(cos(x)) dx (π/2) * ln2(2) * (π/4) * π2/6 π/2 * ln2(2) * π3/24 π * ln2(2) * π3/24 π * ln2(2) * π3/24 (π/2 * ln2(2) * π3/24)

Conclusion

The integral ∫?^(π/2) ln2(cos(x)) dx evaluates to (π/2 * ln2(2) * π3/24), which simplifies to (π/24 * ln2(2) * π3).

This detailed approach showcases the power of combining trigonometric identities, symmetry properties, and Fourier series to solve complex integrals. The final result provides a solid foundation for further exploration into advanced integral calculus techniques.