Question

$Evaluate \int_0^{\pi/2} \frac{1 - \cot x}{\csc x + \cos x} \, dx:$

• A. 0
• B. $\frac{\pi}{4}$
• C. $\infty$
• D. $\frac{\pi}{12}$

Answer 1

Answer: (A)

0

Answer 2

The given integral is:

$I = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cot x}{\csc x + \cos x} dx.$

Analyze the symmetry of the integral. The limits of the integral are symmetric about $\frac{\pi}{4}$, and the integrand contains terms that involve trigonometric functions $\sin x$, $\cos x$, $\cot x$, and $\csc x$. Specifically, consider the property:

$f(x) = -f\left(\frac{\pi}{2} - x\right).$

For the integrand:

$f(x) = \frac{1 - \cot x}{\csc x + \cos x}.$

Using the trigonometric substitutions:

$\cot\left(\frac{\pi}{2} - x\right) = \tan x, \quad \csc\left(\frac{\pi}{2} - x\right) = \sec x, \quad \cos\left(\frac{\pi}{2} - x\right) = \sin x,$

we find that the integrand satisfies the property:

$f(x) + f\left(\frac{\pi}{2} - x\right) = 0.$

Since $f(x)$ is odd with respect to $x = \frac{\pi}{4}$, the integral over the symmetric interval $[0, \frac{\pi}{2}]$ evaluates to 0.

Thus:

$I = 0.$