Question

Find: \[ \int \frac{\cos x \, dx}{1 + \cos x + \sin x}. \]

Hint:

To solve integrals involving trigonometric functions, sometimes it is useful to multiply by the conjugate of the denominator to simplify the expression. Remember, $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$, which can help simplify your work.

Answer 1

The integral is solved by first manipulating the denominator: \[ 1 + \cos x + \sin x = (1 + \sin x + \cos x). \] The integrand is multiplied and divided by the conjugate of the denominator: \[ \frac{1 - \sin x}{1 - \sin x}. \] The integral is rewritten as: \[ \int \frac{\cos x (1 - \sin x)}{(1 + \sin x + \cos x)(1 - \sin x)} dx. \] The denominator simplifies to: \[ (1 + \sin x)(1 - \sin x) = 1 - \sin^2 x = \cos^2 x. \] The integral becomes: \[ \int \frac{\cos x (1 - \sin x)}{\cos^2 x} dx. \] Simplifying the integrand yields: \[ \int \frac{1 - \sin x}{\cos x} dx. \] The integral is split into two parts: \[ \int \frac{1}{\cos x} dx - \int \frac{\sin x}{\cos x} dx. \] The first integral is $\int \sec x \, dx = \ln |\sec x + \tan x| + C_1$. The second integral is $\int \tan x \, dx = -\ln |\cos x| + C_2$. The solution is: \[ \ln |\sec x + \tan x| - \ln |\cos x| + C. \] This simplifies to: \[ \ln \left| \frac{\sec x + \tan x}{\cos x} \right| + C. \]