Question

The integral \[ \int \sqrt{1 + \sin x} \, dx \] is equal to:

• A. \( 2\left( -\sin\left( \frac{x}{2} \right) + \cos\left( \frac{x}{2} \right) \right) + C \)
• B. \( 2\left( \sin\left( \frac{x}{2} \right) - \cos\left( \frac{x}{2} \right) \right) + C \)
• C. \( 2\left( \sin\left( \frac{x}{2} \right) + \cos\left( \frac{x}{2} \right) \right) + C \)
• D. \( 2\left( \sin\left( \frac{x}{2} \right) + \cos\left( \frac{x}{2} \right) \right) + C \)

Hint:

To simplify integrals with trigonometric functions, use trigonometric identities to transform the integrand into a more manageable form.

Answer 1

The Correct Option is A

We use the half angle identity for the trigonometric function: \[ 1 + \sin x = 2\cos^2\left(\frac{x}{2}\right). \] Thus, the integral becomes: \[ \int \sqrt{2\cos^2\left(\frac{x}{2}\right)} \, dx = \int 2\cos\left(\frac{x}{2}\right) \, dx. \] Now, use the substitution \( u = \frac{x}{2} \), \( du = \frac{1}{2}dx \): \[ 2 \int \cos(u) \, du = 2\sin(u) + C = 2\left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right) + C. \]