Integral of log sinx with limits 0 to pi/2

We are tasked with finding the integral of \( \log(\sin x) \) from 0 to \( \frac{\pi}{2} \). The integral is expressed as:

\[
I = \int_0^{\frac{\pi}{2}} \log(\sin x) \, dx
\]

### Step 1: Symmetry property of integrals

There is a known symmetry property involving sine and cosine, which is useful here. It states that:

\[
\int_0^{\frac{\pi}{2}} f(x) \, dx = \int_0^{\frac{\pi}{2}} f\left(\frac{\pi}{2} - x\right) \, dx
\]

Applying this to \( \log(\sin x) \), we substitute \( f(x) = \log(\sin x) \):

\[
I = \int_0^{\frac{\pi}{2}} \log(\sin\left(\frac{\pi}{2} - x\right)) \, dx
\]

But we know that \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \), so the integral becomes:

\[
I = \int_0^{\frac{\pi}{2}} \log(\cos x) \, dx
\]

### Step 2: Adding the two integrals

Now, let's add the original integral and the transformed one:

\[
I + I = \int_0^{\frac{\pi}{2}} \log(\sin x) \, dx + \int_0^{\frac{\pi}{2}} \log(\cos x) \, dx
\]

This simplifies to:

\[
2I = \int_0^{\frac{\pi}{2}} \log(\sin x \cos x) \, dx
\]

### Step 3: Simplifying the integrand

Using the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \), the integral becomes:

\[
2I = \int_0^{\frac{\pi}{2}} \log\left(\frac{1}{2} \sin(2x)\right) \, dx
\]

This can be split into two parts:

\[
2I = \int_0^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right) \, dx + \int_0^{\frac{\pi}{2}} \log(\sin(2x)) \, dx
\]

The first part evaluates as:

\[
\int_0^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right) \, dx = \log\left(\frac{1}{2}\right) \cdot \frac{\pi}{2} = -\frac{\pi}{2} \log(2)
\]

For the second part, let’s change the variable \( u = 2x \), so \( du = 2dx \) and the limits of integration change accordingly:

\[
\int_0^{\frac{\pi}{2}} \log(\sin(2x)) \, dx = \frac{1}{2} \int_0^\pi \log(\sin u) \, du
\]

It's a known result that:

\[
\int_0^\pi \log(\sin u) \, du = -\pi \log(2)
\]

Thus, the second part evaluates as:

\[
\frac{1}{2} (-\pi \log(2)) = -\frac{\pi}{2} \log(2)
\]

### Step 4: Final result

Putting everything together:

\[
2I = -\frac{\pi}{2} \log(2) - \frac{\pi}{2} \log(2)
\]

\[
2I = -\pi \log(2)
\]

Therefore:

\[
I = -\frac{\pi}{2} \log(2)
\]

So, the integral is:

\[
\boxed{I = -\frac{\pi}{2} \log(2)}
\]

---

Would you like any further details or have any questions?

Here are five related questions:
1. What is the integral of \( \log(\cos x) \) from 0 to \( \frac{\pi}{2} \)?
2. How do symmetry properties help in evaluating definite integrals?
3. What is the significance of the identity \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \)?
4. Can you apply this method to evaluate the integral \( \int_0^{\frac{\pi}{2}} \sin^n(x) \, dx \)?
5. What is the result of \( \int_0^\pi \log(\sin x) \, dx \)?

**Tip**: Always look for symmetry and known identities in trigonometric integrals to simplify your work!

Learn how to evaluate the integral of log(sin x) with limits from 0 to pi/2 using symmetry and trigonometric identities. This advanced calculus problem uses known integral results and explores key concepts of definite integrals and symmetry. Suitable for college-level students.

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