Evaluate the integral: I = integral dx/(1 - cos(x))

Evaluate the integral: $$ I = \int \frac{dx}{1 - \cos(x)}. $$

Step 1 · Apply the half-angle identity

To solve this integral, we start by simplifying the integrand. The denominator contains $1-\cos(x)$, which suggests using a trigonometric identity. We use the following identity:

$$ 1-\cos(x) = 2\sin^2\left(\frac{x}{2}\right). $$

Thus, the integral becomes:

$$ I = \int \frac{dx}{2\sin^2\left(\frac{x}{2}\right)}. $$

We can factor out the constant $\dfrac12$ from the integral:

$$ I = \frac{1}{2}\int \frac{dx}{\sin^2\left(\frac{x}{2}\right)}. $$

Next, we recognize that $\dfrac{1}{\sin^2\left(\frac{x}{2}\right)}=\csc^2\left(\dfrac{x}{2}\right)$, so the integral becomes:

$$ I = \frac{1}{2}\int \csc^2\left(\frac{x}{2}\right)\,dx. $$

Step 2 · Substitute

To simplify further, we make a substitution. Let:

$$ u = \frac{x}{2}. $$

Then:

$$ du = \frac{dx}{2}, \quad dx = 2\,du. $$

Step 3 · Transform and integrate

Substituting into the integral:

$$ I = \frac{1}{2}\int \csc^2(u)\cdot 2\,du = \int \csc^2(u)\,du. $$

The integral of $\csc^2(u)$ is a standard result:

$$ \int \csc^2(u)\,du = -\cot(u). $$

Thus, the integral becomes:

$$ I = -\cot(u) + C, $$

where $C$ is the constant of integration.

Step 4 · Substitute back

Finally, substituting back $u = \dfrac{x}{2}$, we obtain the solution:

$$ I = -\cot\left(\frac{x}{2}\right) + C. $$

Summary Table

Additional Notes

• The domain of $I$ excludes $x=2k\pi$ for integer $k$, where $1-\cos x=0$.
• Compare with Problem 57: the identity swaps $\cos^2\to\sin^2$, and correspondingly the antiderivative swaps $\tan\to-\cot$ — a useful pattern to remember for both half-angle forms.
• The result can be checked by differentiating $-\cot(x/2)$ with the chain rule and confirming it returns $\dfrac{1}{1-\cos x}$.