Question

$\int \sqrt{1 + \cos x}$ dx equals

• A. $2\sqrt{2}\sin{\frac{x}{2}} + c$
• B. $-2\sqrt{2}\sin{\frac{x}{2}} + c$
• C. $-2\sqrt{2}\cos{\frac{x}{2}} + c$
• D. $2\sqrt{2}\cos{\frac{x}{2}} + c$

Answer 1

Answer: (A)

$2\sqrt{2}\sin{\frac{x}{2}} + c$

Answer 2

To evaluate the integral $\int \sqrt{1 + \cos x} \, dx$, we use a fundamental trigonometric identity.

Step 1: Apply the trigonometric identity.

We know that $1 + \cos x = 2\cos^2 \frac{x}{2}$. Substitute this into the integral: $\int \sqrt{1 + \cos x} \, dx = \int \sqrt{2\cos^2 \frac{x}{2}} \, dx$

Step 2: Simplify the expression under the integral. $\sqrt{2\cos^2 \frac{x}{2}} = \sqrt{2} \sqrt{\cos^2 \frac{x}{2}} = \sqrt{2} \left| \cos \frac{x}{2} \right|$ For indefinite integrals, without a specified interval for $x$, it is a common convention in multiple-choice questions to assume the principal value of the square root, meaning we consider the interval where $\cos \frac{x}{2}$ is non-negative. For example, if $x \in [-\pi, \pi]$, then $\frac{x}{2} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, where $\cos \frac{x}{2} \ge 0$. Under this assumption, $\left| \cos \frac{x}{2} \right| = \cos \frac{x}{2}$. So the integral becomes: $\int \sqrt{2} \cos \frac{x}{2} \, dx$

Step 3: Perform the integration.

Let $u = \frac{x}{2}$. Then, $du = \frac{1}{2} dx$, which implies $dx = 2 du$. Substitute $u$ and $dx$ into the integral: $\int \sqrt{2} \cos u (2 du) = 2\sqrt{2} \int \cos u \, du$ The integral of $\cos u$ with respect to $u$ is $\sin u$. $2\sqrt{2} \sin u + C$

Step 4: Substitute back for $u$.

Replace $u$ with $\frac{x}{2}$: $2\sqrt{2} \sin \frac{x}{2} + C$

Comparing this result with the given options, we find that it matches option A.