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

The key idea is U Substitution: let $u = 1 + \cos x$, because its derivative $-\sin x$ appears in the numerator.

Step 1: Set $u = 1 + \cos x$. Then $du = -\sin x \, dx$, so $\sin x \, dx = -du$.

Step 2: Substitute into the integral:

Step 3: Integrate: …

Why substitution works here

When you see a fraction where the numerator looks like the derivative of the denominator (or close to it), substitution is your best friend. Here, the denominator is $1 + \cos x$, and its derivative is $-\sin x$. The numerator is $\sin x$ — almost a perfect match, just missing a minus sign. That’s the signal: set $u$ equal to the denominator, and the rest will fall into place.

Step-by-step solution

1. Choose the substitution.

   Let $u = 1 + \cos x$.

   Why this? Because the denominator is $u$, and the derivative of $u$ will simplify the numerator.

2. Differentiate to find $du$.

This means $\sin x \, dx = -du$. The integral’s numerator $\sin x \, dx$ is exactly $-du$.

3. Rewrite the integral in terms of $u$.

4. Integrate with respect to $u$. The integral $\int \frac{du}{u}$ is $\log|u| + C$, so:

5. Substitute back for $x$. Since $u = 1 + \cos x$, we get: $-\log|1 + \cos x| + C$ …

Method: Substitute the denominator to reach a logarithm

Use this when the numerator is (a constant times) the derivative of the denominator, e.g. $\dfrac{\sin x}{1+\cos x}$ where $\dfrac{d}{dx}(1+\cos x)=-\sin x$.

Steps

Step 1: Differentiate the denominator.

$\dfrac{d}{dx}(1+\cos x)=-\sin x$; the numerator $\sin x$ is $-1$ times it.

Step 2: Substitute $u=1+\cos x$. …

Common Mistakes

Mistake 1: Losing the minus sign from $du=-\sin x\,dx$.

Why it's wrong: with $u=1+\cos x$, $\sin x\,dx=-du$, so $\int\dfrac{\sin x}{1+\cos x}\,dx=-\log|1+\cos x|$. Missing the sign gives the wrong answer. Correct approach: substitute the differential exactly, sign included.

Mistake 2: Trying a half-angle expansion unnecessarily. …