Evaluate the integral: I = integral cot x dx

Evaluate the integral: $$ I = \int \cot x \, dx. $$

Explanation of Method

Writing $\cot x = \dfrac{\cos x}{\sin x}$ exposes the numerator-equals-derivative-of-denominator pattern directly: the derivative of $\sin x$ is $\cos x$, so substituting $u=\sin x$ reduces the integral to a logarithmic form with no extra sign to track.

Rewrite $\cot x$ as $\dfrac{\cos x}{\sin x}$.
Substitute $u=\sin x$.
Integrate using the standard logarithmic antiderivative.
Substitute back and add the constant of integration.

Step 1 · Rewrite in terms of sine and cosine

We recall that the cotangent function can be rewritten as:

$$ \cot x = \frac{\cos x}{\sin x}. $$

Step 2 · Substitution

Using the substitution:

$$ u = \sin x, \quad du = \cos x\,dx, $$

we rewrite the integral as:

$$ I = \int \frac{du}{u}. $$

Step 3 · Integrate

The integral of $\dfrac{1}{u}$ is:

$$ I = \ln|u| + C. $$

Step 4 · Substitute back

Substituting back $u = \sin x$:

$$ I = \ln|\sin x| + C. $$

Summary Table

Quantity	Value
Substitution	$u = \sin x$
$\cos x\,dx$ in terms of $du$	$du$
Antiderivative	$\ln\|u\|$

Additional Notes

The domain of $I$ excludes $x = k\pi$ for integer $k$, where $\sin x = 0$.
Compare with Problem 93: $\int \tan x\,dx$ picks up a minus sign because $\cos x$'s derivative is $-\sin x$, while $\int \cot x\,dx$ does not, because $\sin x$'s derivative is $+\cos x$.
The result can be checked by differentiating $\ln|\sin x|$ with the chain rule and confirming it returns $\cot x$.

Result

The solution to the integral is:

$$ I = \ln|\sin x| + C. $$