Q.$\int_{0}^{\pi} x\log\sin x\,dx$

The key idea is to use the property of definite integrals with the substitution $x \to \pi - x$, exploiting the symmetry of $\sin x$ over $[0, \pi]$.

Let $I = \int_{0}^{\pi} x \log \sin x \, dx$.

Substitute $x = \pi - t$, so $dx = -dt$. When $x = 0$, $t = \pi$; when $x = \pi$, $t = 0$. Then:

Since $\sin(\pi - t) = \sin t$, and renaming $t$ back to $x$, we have:

Add the two expressions for $I$:

$2I = \int_{0}^{\pi} [x + (\pi - x)] \log \sin x \, dx = \pi \int_{0}^{\pi} \log \sin x \, dx.$ …

The problem asks for $I = \int_0^\pi x \log\sin x \, dx$. The presence of $x$ multiplied by a function symmetric about $\pi/2$ suggests using the symmetry property of definite integrals. For any function $f(x)$ integrable on $[0,a]$, we have $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$. Here $a = \pi$, so we can replace $x$ by $\pi - x$ in the integral.

Let’s work through it step by step.

1. Apply the substitution $x \to \pi - x$. Let $I = \int_0^\pi x \log\sin x \, dx$. Using $x = \pi - t$, when $x=0$, $t=\pi$; when $x=\pi$, $t=0$. The integral becomes

Since $\sin(\pi - t) = \sin t$, we have

2. Add the two expressions for $I$. We now have two forms:

Adding them:

So

3. Evaluate $J = \int_0^\pi \log\sin x \, dx$. This is a classic integral. Use symmetry again:

because $\sin x$ is symmetric about $\pi/2$ on $[0,\pi]$.

Now consider $K = \int_0^{\pi/2} \log\sin x \, dx$. A standard trick: substitute $x \to \pi/2 - x$ to get $K = \int_0^{\pi/2} \log\cos x \, dx$. Adding:

So

Let $u = 2x$, then $dx = du/2$, limits $0$ to $\pi$: …

Method: King property with a symmetric log integrand

For $\int_0^a x\,f(x)\,dx$ where $f(a-x)=f(x)$, the property $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ removes the $x$; a known "log-sine" value finishes it.

Steps

Step 1: Reflect and add.

$I=\int_0^a x f(x)\,dx=\int_0^a (a-x)f(x)\,dx$, so $2I=a\int_0^a f(x)\,dx$.

Step 2: Reduce the log-sine integral. …

Common Mistakes

Mistake 1: Not knowing (or deriving) $\int_0^\pi\log\sin x\,dx=-\pi\log 2$.

Why it's wrong: without this value the problem cannot be finished. Correct approach: quote it, or derive it via the $\sin 2x=2\sin x\cos x$ doubling identity.

Mistake 2: Sign confusion under $x\to\pi-x$. …