Question

Evaluate the integral: \[ \int_0^{\frac{\pi}{2}} \frac{dx}{1 + \cos x} \]

• A. -2
• B. 2
• C. 1
• D. -1

Hint:

For integrals involving \(1 + \cos x\), use the half angle identity to simplify the expression.

Answer 1

The Correct Option is C

Step 1: Simplify the integrand using a trigonometric identity.
We use the identity \[ 1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right) \] Thus, the integral becomes \[ \int_0^{\frac{\pi}{2}} \frac{dx}{2 \cos^2 \left(\frac{x}{2}\right)} \] \[ = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sec^2 \left(\frac{x}{2}\right) dx \] Step 2: Substitute and integrate.
Let \( u = \frac{x}{2} \), hence \( du = \frac{dx}{2} \). The limits change accordingly: when \(x = 0\), \(u = 0\); and when \(x = \frac{\pi}{2}\), \(u = \frac{\pi}{4}\). Therefore, the integral becomes \[ \frac{1}{2} \times 2 \int_0^{\frac{\pi}{4}} \sec^2 u \, du \] \[ = \int_0^{\frac{\pi}{4}} \sec^2 u \, du \] Step 3: Evaluate the integral.
The integral of \(\sec^2 u\) is \(\tan u\), so \[ \int_0^{\frac{\pi}{4}} \sec^2 u \, du = \tan \left(\frac{\pi}{4}\right) - \tan(0) \] \[ = 1 - 0 = 1 \]