Evaluate the integral: I = integral tan x dx

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

Explanation of Method

Writing $\tan x = \dfrac{\sin x}{\cos x}$ exposes a numerator-equals-derivative-of-denominator pattern (up to sign): the derivative of $\cos x$ is $-\sin x$, so substituting $u=\cos x$ reduces the integral directly to a logarithmic form.

Rewrite $\tan x$ as $\dfrac{\sin x}{\cos x}$.
Substitute $u=\cos x$.
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 tangent function can be rewritten as:

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

Step 2 · Substitution

Using the substitution:

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

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 $u = \cos x$:

$$ I = -\ln|\cos x| + C. $$

Summary Table

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

Additional Notes

The domain of $I$ excludes $x = \tfrac\pi2+k\pi$ for integer $k$, where $\cos x = 0$.
An equivalent common form of the answer is $\ln|\sec x|+C$, since $-\ln|\cos x| = \ln|1/\cos x| = \ln|\sec x|$.
The result can be checked by differentiating $-\ln|\cos x|$ with the chain rule and confirming it returns $\tan x$.

Result

The solution to the integral is:

$$ I = -\ln|\cos x| + C. $$