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

Concept: U Substitution — rewrite the integrand using a trigonometric identity to simplify.

First, use the identity $\cos x = 2\cos^2\frac{x}{2} - 1 = 1 - 2\sin^2\frac{x}{2}$. Then:

Now integrate:

Let $u = \frac{x}{2}$, so $dx = 2\,du$. Then: …

Why this approach works

When you see a ratio of $1 \pm \cos x$, your first instinct should be half-angle formulas. They exist precisely to simplify such expressions. The identity $\cos x = 2\cos^2\frac{x}{2} - 1 = 1 - 2\sin^2\frac{x}{2}$ lets us rewrite both numerator and denominator in terms of $\frac{x}{2}$, and the ratio collapses beautifully into a single squared tangent.

Why tangent? Because $\frac{1 - \cos x}{1 + \cos x}$ is a classic form for $\tan^2\frac{x}{2}$. Once you have $\tan^2 u$, you integrate it by recalling that $\tan^2 u = \sec^2 u - 1$ — and $\sec^2 u$ integrates to $\tan u$, while $1$ integrates to $u$. The whole thing becomes a clean, two-step process.

Step-by-step solution

1. Apply the half-angle identity.

We use $\cos x = 1 - 2\sin^2\frac{x}{2} = 2\cos^2\frac{x}{2} - 1$. Then:

So the integrand becomes:

2. Set up the integral.

We now have:

3. Substitute to simplify.

Let $u = \frac{x}{2}$, so $x = 2u$ and $dx = 2\,du$. Then:

$\int \tan^2\frac{x}{2} \, dx = \int \tan^2 u \cdot 2\,du = 2\int \tan^2 u \, du$ …

Method: Half-angle identities to simplify $\dfrac{1\pm\cos x}{1\pm\cos x}$ forms

A ratio built from $1-\cos x$ and $1+\cos x$ collapses to a single squared trig function via half-angle identities; then integrate that.

Steps

Step 1: Rewrite numerator and denominator with half-angle forms.

so the ratio becomes $\dfrac{2\sin^2(x/2)}{2\cos^2(x/2)}=\tan^2\tfrac{x}{2}$.

Step 2: Convert $\tan^2$ using the Pythagorean identity.

turns the integrand into a standard $\sec^2$ minus a constant. …

Common Mistakes

Mistake 1: Forgetting the factor $2$ from $dx=2\,du$ when substituting $u=x/2$.

Why it's wrong: integrating $\sec^2(x/2)$ gives $2\tan(x/2)$, not $\tan(x/2)$, because the argument's slope is $\tfrac12$. Correct approach: $\displaystyle\int\sec^2\tfrac{x}{2}\,dx=2\tan\tfrac{x}{2}$, yielding $2\tan\tfrac{x}{2}-x+C$.

Mistake 2: Cancelling $\dfrac{1-\cos x}{1+\cos x}$ to a wrong simplification. …