Evaluate the integral: I = integral x/(cos²x) dx

Evaluate the integral: $$ I = \int \frac{x}{\cos^2x}\,dx. $$

Explanation of Method

Rewriting $1/\cos^2x=\sec^2x$ and applying integration by parts with $u=x$, $dv=\sec^2x\,dx$ leaves a leftover integral of $\tan x$ alone, which is a standard logarithmic result.

Choose $u=x$, $dv=\sec^2x\,dx$.
Differentiate $u$ and integrate $dv$.
Apply the integration by parts formula and integrate the remainder.

Step 1 · Set up integration by parts

Let:

$$ u = x \quad \Rightarrow \quad du = dx, \qquad dv = \sec^2x\,dx \quad \Rightarrow \quad v = \tan x. $$

Step 2 · Apply the formula

$$ I = x\tan x - \int \tan x\,dx. $$

Step 3 · Integrate

Recall the standard result $\int \tan x\,dx = -\ln|\cos x|$:

$$ I = x\tan x - \left(-\ln|\cos x|\right) + C = x\tan x + \ln|\cos x| + C. $$

Summary Table

Quantity	Value
$u$	$x$
$dv$	$\sec^2x\,dx$
Antiderivative	$x\tan x+\ln\|\cos x\|$

Additional Notes

The domain excludes $x=\tfrac\pi2+k\pi$ for integer $k$, where $\cos x=0$.
A common sign slip here is recalling $\int\tan x\,dx=\ln|\cos x|$ instead of the correct $-\ln|\cos x|$; differentiating $\ln|\cos x|$ gives $-\tan x$, the negative of the integrand, which is a quick way to catch the error.
The result can be checked by differentiating $x\tan x+\ln|\cos x|$: this gives $\tan x+x\sec^2x-\tan x=x\sec^2x=x/\cos^2x$, matching the integrand.

Result

The solution to the integral is:

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