Integral Calculation - Exercise 27

We are asked to evaluate the following indefinite integral:

$$ \int 2 \cdot \tan^2(x) - 1 \ dx $$

This integral can be tackled using different techniques.

Solution 1

Let’s begin by breaking the expression apart using the linearity of integration:

$$ \int 2 \cdot \tan^2(x) \ dx - \int 1 \ dx $$

$$ = 2 \cdot \int \tan^2(x) \ dx - \int 1 \ dx $$

The second term is immediate: \( \int 1 \ dx = x + c \).

So we have:

$$ 2 \cdot \int \tan^2(x) \ dx - x + c $$

We now recall the trigonometric identity:

$$ \tan^2(x) = \frac{1}{\cos^2(x)} - 1 $$

Explanation. Starting from the Pythagorean identity: $$ \sin^2(x) + \cos^2(x) = 1 $$ Dividing both sides by \( \cos^2(x) \): $$ \frac{\sin^2(x)}{\cos^2(x)} + 1 = \frac{1}{\cos^2(x)} $$ Hence, $$ \frac{\sin^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} - 1 $$ By the definition of tangent: $$ \tan^2(x) = \frac{1}{\cos^2(x)} - 1 $$

Substitute this into the integral:

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

Apply linearity again:

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

Now evaluate each term. We know: \( \int \frac{1}{\cos^2(x)} \ dx = \tan(x) + C \), and \( \int 1 \ dx = x + C \)

Therefore:

$$ 2 \cdot (\tan(x) - x) - x + c $$

$$ = 2 \tan(x) - 2x - x + c $$

$$ = 2 \tan(x) - 3x + c $$

This is the final result.

Solution 2

Let’s start again from the original integral:

$$ \int 2 \cdot \tan^2(x) - 1 \ dx $$

We use the definition of tangent: \( \tan(x) = \frac{\sin(x)}{\cos(x)} \)

So we rewrite the integrand:

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

Which simplifies to:

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

Now apply the identity \( \sin^2(x) = 1 - \cos^2(x) \), and substitute:

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

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

$$ = \int \frac{2 - 3\cos^2(x)}{\cos^2(x)} \ dx $$

Now split the fraction:

$$ = \int \left( \frac{2}{\cos^2(x)} - \frac{3\cos^2(x)}{\cos^2(x)} \right) dx $$

$$ = \int \frac{2}{\cos^2(x)} \ dx - \int 3 \ dx $$

Apply linearity once more:

$$ 2 \cdot \int \frac{1}{\cos^2(x)} \ dx - 3 \cdot \int 1 \ dx $$

Both integrals are standard: \( \int \frac{1}{\cos^2(x)} \ dx = \tan(x) + C \), \( \int 1 \ dx = x + C \)

Putting it all together:

$$ 2 \tan(x) - 3x + c $$

Once again, we arrive at the same result.

And so on...