Question

The indefinite integral $\int \frac{1}{1 + \cos x} \, dx$ is equal to:

• A. $\frac{1}{2}\tan\frac{x}{2} + C$
• B. $-\frac{1}{2}\cot\frac{x}{2} + C$
• C. $-\cot\frac{x}{2} + C$
• D. $\tan\frac{x}{2} + C$

Hint:

Alternatively, you can rationalize the denominator by multiplying the numerator and denominator by $(1 - \cos x)$: \[ \frac{1}{1+\cos x} \cdot \frac{1-\cos x}{1-\cos x} = \frac{1-\cos x}{1-\cos^2 x} = \frac{1-\cos x}{\sin^2 x} = \csc^2 x - \csc x \cot x \] Integrating this gives $-\cot x + \csc x + C$, which simplifies exactly to $\tan\left(\frac{x}{2}\right) + C$ using half-angle properties!

Answer 1

The Correct Option is D

Concept: To solve an indefinite integral involving trigonometric functions, we try to simplify the integrand using standard trigonometric identities. A very helpful half-angle identity for $1 + \cos x$ is: \[ 1 + \cos x = 2\cos^2\left(\frac{x}{2}\right) \] Using this substitution transforms the denominator into a single squared trigonometric function, which can then be written as a standard integrable function like $\sec^2\left(\frac{x}{2}\right)$.

Step 1: Apply the trigonometric identity to the denominator.
Let the given integral be denoted by $I$: \[ I = \int \frac{1}{1 + \cos x} \, dx \] We substitute the identity $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$ directly into the integral: \[ I = \int \frac{1}{2\cos^2\left(\frac{x}{2}\right)} \, dx \]

Step 2: Simplify using reciprocal trigonometric identities.
We can pull the constant factor $\frac{1}{2}$ out of the integration sign: \[ I = \frac{1}{2} \int \frac{1}{\cos^2\left(\frac{x}{2}\right)} \, dx \] Since the reciprocal of $\cos\theta$ is $\sec\theta$, we know that $\frac{1}{\cos^2\left(\frac{x}{2}\right)} = \sec^2\left(\frac{x}{2}\right)$. Substituting this gives: \[ I = \frac{1}{2} \int \sec^2\left(\frac{x}{2}\right) \, dx \]

Step 3: Integrate using substitution or standard formulae.
We know from standard integration rules that the integral of $\sec^2(kx)$ is $\frac{\tan(kx)}{k}$. Here, $k = \frac{1}{1/2}$ inside the linear term argument: Let $u = \frac{x}{2}$, then $du = \frac{1}{2} dx$, which means $2 du = dx$. Substituting these values back into the expression: \[ I = \frac{1}{2} \int \sec^2(u) \cdot (2 du) = \int \sec^2(u) \, du \] The standard derivative of $\tan(u)$ is $\sec^2(u)$, so: \[ I = \tan(u) + C \] Substituting back $u = \frac{x}{2}$: \[ I = \tan\left(\frac{x}{2}\right) + C \]