Q.Integrate the following function: $\frac{\sin^2 x}{1 + \cos x}$

The key idea is to use the identity $\sin^2 x = 1 - \cos^2 x$, which factors as a difference of squares and cancels with the denominator.

First, rewrite the numerator:

Factor the numerator:

$\frac{(1 - \cos x)(1 + \cos x)}{1 + \cos x}.$ …

When you see a rational expression involving $\sin^2 x$ and $1 + \cos x$, your first instinct might be to try a substitution like $t = \tan(x/2)$ — the universal trigonometric substitution. But that’s overkill here. The smarter path is to notice that $\sin^2 x$ and $\cos x$ are related through the Pythagorean identity: $\sin^2 x = 1 - \cos^2 x$. This lets you factor the numerator as a difference of squares, which cancels beautifully with the denominator.

Let’s walk through it.

1. Rewrite the numerator Use $\sin^2 x = 1 - \cos^2 x$. The integrand becomes:

2. Factor the numerator $1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$. So:

3. Cancel the common factor Provided $1 + \cos x \neq 0$ (which is true except at isolated points where $\cos x = -1$, and those don’t affect the indefinite integral), we get: $1 - \cos x$ …

Method: Simplify a trig fraction with the Pythagorean identity before integrating

When a squared sine or cosine sits over a $1\pm\cos x$ (or $1\pm\sin x$) denominator, factor using $\sin^2 x=1-\cos^2 x=(1-\cos x)(1+\cos x)$ so the denominator cancels.

Steps

Step 1: Rewrite the squared term as a difference of squares.

Step 2: Cancel the common factor with the denominator.

$\frac{\sin^2 x}{1+\cos x}=\frac{(1-\cos x)(1+\cos x)}{1+\cos x}=1-\cos x$ …

Common Mistakes

Mistake 1: Reaching straight for a half-angle substitution.

Why it's wrong: it works but is far longer than needed; the integrand simplifies algebraically first. Correct approach: use $\sin^2 x=(1-\cos x)(1+\cos x)$ so the $1+\cos x$ cancels, leaving $1-\cos x$.

Mistake 2: Cancelling incorrectly and losing the constant term. …