Integral Calculation Exercise 22

We are asked to evaluate the integral

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

To approach this, we apply the substitution method.

We begin by observing that the derivative of $ \sin x $ is:

$$ d(\sin x) = \cos x \ dx $$

Solving for $ dx $, we get:

$$ dx = \frac{d(\sin x)}{\cos x} $$

Substituting into the original integral yields:

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

The cosine terms cancel out, leaving:

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

We now use the Pythagorean identity $ \sin^2 x + \cos^2 x = 1 $, which implies:

$ \cos^2 x = 1 - \sin^2 x $

Substituting this into the integrand:

$$ \int \frac{1}{2 - (1 - \sin^2 x)} \ d(\sin x) $$

Simplifying the denominator:

$$ \int \frac{1}{1 + \sin^2 x} \ d(\sin x) $$

Letting $ t = \sin x $, we have:

$$ \int \frac{1}{1 + t^2} \ dt $$

This is a standard integral, and we know that:

$ \int \frac{1}{1 + t^2} \ dt = \arctan(t) + c $

Substituting back $ t = \sin x $, we obtain:

$$ \arctan(\sin x) + c $$

Hence, the solution to the integral is:

$$ \int \frac{\cos x}{2 - \cos^2 x} \ dx = \arctan(\sin x) + c $$

This completes the evaluation.