Question

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

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

Hint:

When analyzing integrals with symmetric limits, it's often helpful to check the symmetry of the integrand. If the integrand has an odd symmetry with respect to the midpoint of the interval, the integral over that symmetric range will evaluate to zero. For example, using properties like \( f(x) = -f\left(\frac{\pi}{2} - x\right) \) can simplify the process of evaluating such integrals by showing that the positive and negative parts of the integrand cancel each other out.

Answer 1

The Correct Option is A

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:

\[ I = 0. \]

Answer 2

To evaluate the integral \( \int_0^{\pi/2} \frac{1 - \cot x}{\csc x + \cos x} \, dx \), we begin by simplifying the integrand.

First, recall the trigonometric identities: 
\(\cot x = \frac{\cos x}{\sin x}\) and \(\csc x = \frac{1}{\sin x}\).

Rewrite the integrand:

\[\frac{1 - \cot x}{\csc x + \cos x} = \frac{1 - \frac{\cos x}{\sin x}}{\frac{1}{\sin x} + \cos x}.\]

Simplify both the numerator and the denominator:

The numerator becomes:

\[1 - \frac{\cos x}{\sin x} = \frac{\sin x - \cos x}{\sin x}.\]

The denominator becomes:

\[\frac{1}{\sin x} + \cos x = \frac{1 + \cos x \sin x}{\sin x}.\]

Thus, the integrand simplifies to:

\[\frac{\frac{\sin x - \cos x}{\sin x}}{\frac{1 + \cos x \sin x}{\sin x}} = \frac{\sin x - \cos x}{1 + \cos x \sin x}.\]

Let \( I = \int_0^{\pi/2} \frac{\sin x - \cos x}{1 + \cos x \sin x} \, dx.\)

Consider the substitution \( x = \frac{\pi}{2} - t \). Then \( dx = -dt \), and the limits change from \( x = 0 \) to \( t = \frac{\pi}{2} \), and \( x = \frac{\pi}{2} \) to \( t = 0 \).

The integral becomes:

\[I = \int_{\pi/2}^{0} \frac{\sin(\frac{\pi}{2} - t) - \cos(\frac{\pi}{2} - t)}{1 + \cos(\frac{\pi}{2} - t)\sin(\frac{\pi}{2} - t)} \, (-dt).\]

Using the identities \(\sin(\frac{\pi}{2} - t) = \cos t\) and \(\cos(\frac{\pi}{2} - t) = \sin t\), we have:

\[I = \int_0^{\pi/2} \frac{\cos t - \sin t}{1 + \sin t \cos t} \, dt = \int_0^{\pi/2} -\frac{\sin t - \cos t}{1 + \cos t \sin t} \, dt.\]

Thus, we find:

\[I + I = 0,\]

which implies:

\[2I = 0 \Rightarrow I = 0.\]

Therefore, the value of the integral is \({0}\).