Integral Calculation Exercise 29

We’re asked to evaluate the following integral:

$$ \int \frac{( \sin x - \cos x)^2}{\cos^2 x} \ dx $$

We begin by expanding the square in the numerator:

$$ \int \frac{\sin^2 x - 2 \sin x \cos x + \cos^2 x}{\cos^2 x} \ dx $$

Recalling the Pythagorean identity, \( \sin^2 x + \cos^2 x = 1 \), the integrand simplifies to:

$$ \int \frac{1 - 2 \sin x \cos x}{\cos^2 x} \ dx $$

We now split the expression into two separate terms:

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

$$ = \int \frac{1}{\cos^2 x} \ dx - \int \frac{2 \sin x}{\cos x} \ dx $$

Applying linearity:

$$ = \int \sec^2 x \ dx - 2 \int \tan x \ dx $$

The first integral is standard:

$$ \int \sec^2 x \ dx = \tan x + C $$

So we have:

$$ \tan x + C - 2 \int \tan x \ dx $$

To compute the second integral, we let t = \cos x and apply substitution:

$$ t = \cos x \quad \Rightarrow \quad dt = -\sin x \ dx $$

Solving for \( dx \):

$$ dx = - \frac{1}{\sin x} \ dt $$

Substituting into the integral:

$$ -2 \int \frac{\sin x}{\cos x} \cdot \frac{1}{\sin x} \cdot (-1) \ dt = 2 \int \frac{1}{\cos x} \ dt $$

That gives us:

$$ \tan x + C + 2 \int \sec x \ dt $$

Now, this step reveals a subtle issue: the variable of integration has been changed to \( t \), but the integrand is still expressed in terms of \( x \). So, strictly speaking, substitution here is unnecessary. A more direct approach is simply to integrate \( \tan x \):

$$ \int \tan x \ dx = -\log |\cos x| + C $$

Thus, we find:

$$ \tan x - 2(-\log |\cos x|) + C = \tan x + 2 \log |\cos x| + C $$

Therefore, the final result is:

$$ \int \frac{( \sin x - \cos x)^2}{\cos^2 x} \ dx = \tan x + 2 \log |\cos x| + C $$

And that completes the solution.