Q.Integrate the following function: $\frac{1}{1 + \cot x}$

Concept: U Substitution — rewrite $\cot x$ in terms of $\sin x$ and $\cos x$, then simplify.

First, note that

Now write the numerator as a combination of the derivative of the denominator:

Thus

$\int \frac{\sin x}{\sin x + \cos x}\,dx = \frac{1}{2}\int dx - \frac{1}{2}\int \frac{\cos x - \sin x}{\sin x + \cos x}\,dx.$ …

We want to integrate $\frac{1}{1 + \cot x}$. At first glance, this looks messy because $\cot x$ is a ratio of trigonometric functions. But the moment you see a rational expression involving $\cot x$, the instinct should be: rewrite everything in terms of $\sin x$ and $\cos x$. That’s the universal first step — it turns an unfamiliar form into something you can manipulate.

1. Rewrite the integrand in terms of sine and cosine

Recall that $\cot x = \frac{\cos x}{\sin x}$. So:

Now the problem becomes:

This is much friendlier. But it’s still not directly integrable by a simple substitution — the denominator is a sum of two different trig functions.

2. Spot the symmetry trick

Here’s the insight: if you also consider the integral of $\frac{\cos x}{\sin x + \cos x}$, something beautiful happens. Let’s define:

Now add them:

That’s one equation linking $I$ and $J$.

3. Subtract them to get a logarithmic form

Now subtract:

This looks like a candidate for substitution. Notice that the derivative of the denominator $\sin x + \cos x$ is $\cos x - \sin x = -(\sin x - \cos x)$. That’s exactly the numerator up to a sign.

Let $u = \sin x + \cos x$. Then $du = (\cos x - \sin x) \, dx = -(\sin x - \cos x) \, dx$. So:

$I - J = \int \frac{\sin x - \cos x}{\sin x + \cos x} \, dx = \int \frac{-du}{u} = -\log|u| + C_2 = -\log|\sin x + \cos x| + C_2.$ …

Method: Split the numerator as (part denominator) + (part its derivative)

Use this for $\dfrac{1}{1+\cot x}$-type integrals: convert to sines/cosines, then write the numerator as $A\,(\text{denominator})+B\,(\text{derivative of denominator})$.

Steps

Step 1: Convert to sine and cosine.

Step 2: Express the numerator in terms of the denominator and its derivative.

Write $\sin x=A(\sin x+\cos x)+B(\cos x-\sin x)$; matching coefficients gives $A=\tfrac12,\ B=-\tfrac12$. …

Common Mistakes

Mistake 1: Leaving the integrand as $\dfrac{1}{1+\cot x}$.

Why it's wrong: it must first be converted to $\dfrac{\sin x}{\sin x+\cos x}$ to be workable. Correct approach: write $\cot x=\dfrac{\cos x}{\sin x}$ and simplify.

Mistake 2: Not using the $A f + B f'$ decomposition. …