Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/2}(2\log\sin x-\log\sin 2x)\,dx$

The key idea is to use the symmetry of the integrand and the standard result $\int_{0}^{\pi/2} \log \sin x \, dx = -\frac{\pi}{2}\log 2$.

First, simplify the integrand:

So the integral becomes:

$I = \int_{0}^{\pi/2} (\log\sin x - \log\cos x) \, dx - \log 2 \int_{0}^{\pi/2} 1 \, dx.$ …

The key to this problem lies in recognising that the integrand can be rewritten using logarithm properties, and then applying the standard definite integral symmetry trick. When you see an integral from $0$ to $\frac{\pi}{2}$ involving $\log\sin x$, your first instinct should be to use the substitution $x \to \frac{\pi}{2} - x$ — this often creates a second copy of the same integral, allowing you to solve for it algebraically.

Let’s break it down.

1. Simplify the integrand using log rules. The expression inside the integral is $2\log\sin x - \log\sin 2x$. Using $\log a^b = b\log a$ and $\log a - \log b = \log\frac{a}{b}$, we get:

Now recall the double-angle identity: $\sin 2x = 2\sin x \cos x$. Substituting:

So the integrand becomes $\log\left(\frac{1}{2}\tan x\right) = \log\frac{1}{2} + \log(\tan x) = -\log 2 + \log(\tan x)$.

Therefore, the integral $I$ is:

The first part is easy: $\int_{0}^{\pi/2} dx = \frac{\pi}{2}$, so that term is $-\frac{\pi}{2}\log 2$.

2. Now handle the tricky part: $J = \int_{0}^{\pi/2} \log(\tan x) \, dx$.

   At first glance, $\log(\tan x)$ looks like it might be messy. But here’s the beautiful symmetry trick: use the substitution $x \to \frac{\pi}{2} - x$.

   Let $x = \frac{\pi}{2} - t$. Then $dx = -dt$, and when $x=0$, $t=\frac{\pi}{2}$; when $x=\frac{\pi}{2}$, $t=0$. So:

Now, $\tan\left(\frac{\pi}{2} - t\right) = \cot t = \frac{1}{\tan t}$. So: …

Method: Simplify logs first, then use the $\int_0^{\pi/2}\log(\tan x)=0$ symmetry

For integrands built from $\log\sin$, $\log\cos$, $\log\sin 2x$, collapse them with log and double-angle rules, then exploit the reflection $x\to\tfrac{\pi}{2}-x$ that sends $\log\tan x\to-\log\tan x$.

Steps

Step 1: Compress with log laws.

Use $a\log m=\log m^a$, $\log m-\log n=\log\tfrac{m}{n}$, and $\sin 2x=2\sin x\cos x$ to reduce the integrand to $\log(\tan x)$ plus a constant.

Step 2: Split off the constant. …

Common Mistakes

Mistake 1: Expanding $\log\sin 2x$ incorrectly.

Why it's wrong: $\sin 2x=2\sin x\cos x$, so $\log\sin 2x=\log 2+\log\sin x+\log\cos x$; dropping a term leaves the wrong constant. Correct approach: apply the product-to-sum log rule fully.

Mistake 2: Not recognising $\int_0^{\pi/2}\log\tan x\,dx=0$. …