Question

\(\displaystyle \int_{0}^{\pi} \frac{\cos x}{\sqrt{1 - \sin^2 x}} \, dx =\)

• A. \( \pi \)
• B. \( \frac{\pi}{2} \)
• C. \( \)
• D. \( 0 \)

Hint:

Use the identity \( \sqrt{1 - \sin^2 x} = |\cos x| \) carefully — absolute values change sign based on interval.

Answer 1

The Correct Option is D

Note that \( \sqrt{1 - \sin^2 x} = \sqrt{\cos^2 x} = |\cos x| \) So the integrand becomes: \[ \frac{\cos x}{|\cos x|} = \begin{cases} 1, & \text{if } \cos x>0 \\ -1, & \text{if } \cos x<0 \end{cases} \] Now analyze the interval \( [0, \pi] \): - In \( (0, \frac{\pi}{2}) \), \( \cos x>0 \) → value is \( +1 \) - In \( (\frac{\pi}{2}, \pi) \), \( \cos x<0 \) → value is \( -1 \) - At \( x = \frac{\pi}{2} \), \( \cos x = 0 \), so integrand undefined, but it's a single point and doesn’t affect definite integral. Now break the integral: \[ \int_0^{\pi} \frac{\cos x}{|\cos x|} dx = \int_0^{\frac{\pi}{2}} 1\,dx + \int_{\frac{\pi}{2}}^{\pi} (-1)\,dx = \frac{\pi}{2} - \frac{\pi}{2} = 0 \]