Q.Integrate the following function: $\cot x \log \sin x$

The key idea is U Substitution: let $u = \log \sin x$, because the derivative of $\log \sin x$ is $\cot x$, which appears as a factor.

Step 1: Let $u = \log \sin x$. Then $du = \frac{\cos x}{\sin x} \, dx = \cot x \, dx$.

Step 2: The integral becomes $\int u \, du$. …

Why substitution works here

When you see a product like $\cot x \cdot \log \sin x$, your first instinct should be to check if one factor is the derivative of the other. Here, $\frac{d}{dx} (\log \sin x) = \frac{\cos x}{\sin x} = \cot x$. That’s a dead giveaway: the integrand is of the form $f(x) \cdot f'(x)$, which integrates to $\frac{1}{2}[f(x)]^2 + C$.

This is the core idea behind the u-substitution we’ll use.

Step-by-step solution

1. Set up the substitution Let $u = \log \sin x$. Then differentiate:

So $du = \cot x \, dx$.

2. Rewrite the integral The original integral is $\int \cot x \log \sin x \, dx$. Substituting $u$ and $du$:

3. Integrate

4. Back-substitute Replace $u$ with $\log \sin x$: $\frac{1}{2} (\log \sin x)^2 + C$ …

Method: Substitute $u=\log\sin x$ when $\cot x\,dx$ is present

Use this when $\log\sin x$ multiplies $\cot x$: note $\cot x$ is exactly the derivative of $\log\sin x$.

Steps

Step 1: Recognise the derivative pairing.

$\dfrac{d}{dx}\log\sin x=\dfrac{\cos x}{\sin x}=\cot x$, which multiplies $\log\sin x$ here.

Step 2: Substitute $u=\log\sin x$.

Then $du=\cot x\,dx$, so the integral becomes $\displaystyle\int u\,du$. …

Common Mistakes

Mistake 1: Not recognising $\cot x$ as the derivative of $\log\sin x$.

Why it's wrong: overlooking $\dfrac{d}{dx}\log\sin x=\cot x$ blocks the substitution. Correct approach: set $u=\log\sin x$, so $du=\cot x\,dx$.

Mistake 2: Confusing this with $\int\cot x\,dx$.

Why it's wrong: the extra $\log\sin x$ factor makes it $\int u\,du$, not $\log|\sin x|$. Correct approach: substitute so the integrand becomes $u\,du$. …