Evaluating the Definite Integral \( \int_{0}^{\frac{\pi}{2}} \frac{1 - \cot x}{\cosec x + \cos x} dx \)

We are asked to evaluate the definite integral:

\( I = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cot x}{\cosec x + \cos x} dx \)

First, let's simplify the integrand using basic trigonometric identities. We know that \( \cot x = \frac{\cos x}{\sin x} \) and \( \cosec x = \frac{1}{\sin x} \). Substitute these into the integrand:

\( \frac{1 - \cot x}{\cosec x + \cos x} = \frac{1 - \frac{\cos x}{\sin x}}{\frac{1}{\sin x} + \cos x} \)

To simplify the complex fraction, find a common denominator in the numerator and the denominator:

\( = \frac{\frac{\sin x - \cos x}{\sin x}}{\frac{1 + \cos x \sin x}{\sin x}} \)

Now, multiply the numerator by the reciprocal of the denominator:

\( = \frac{\sin x - \cos x}{\sin x} \times \frac{\sin x}{1 + \cos x \sin x} \)

Cancel out the \( \sin x \) term:

\( = \frac{\sin x - \cos x}{1 + \sin x \cos x} \)

So the integral becomes:

\( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} dx \)

This form of integral is often solved using a property of definite integrals. The property states that for a function \( f(x) \) and a limit \( a \):

\( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \)

In our case, the upper limit is \( a = \frac{\pi}{2} \). Let \( f(x) = \frac{\sin x - \cos x}{1 + \sin x \cos x} \). We need to find \( f(\frac{\pi}{2} - x) \).

Substitute \( x \) with \( \frac{\pi}{2} - x \) in \( f(x) \):

\( f(\frac{\pi}{2} - x) = \frac{\sin(\frac{\pi}{2} - x) - \cos(\frac{\pi}{2} - x)}{1 + \sin(\frac{\pi}{2} - x) \cos(\frac{\pi}{2} - x)} \)

Using the complementary angle identities, \( \sin(\frac{\pi}{2} - x) = \cos x \) and \( \cos(\frac{\pi}{2} - x) = \sin x \):

\( f(\frac{\pi}{2} - x) = \frac{\cos x - \sin x}{1 + \cos x \sin x} \)

Notice that \( \cos x - \sin x = -(\sin x - \cos x) \). So:

\( f(\frac{\pi}{2} - x) = \frac{-(\sin x - \cos x)}{1 + \sin x \cos x} = - \left( \frac{\sin x - \cos x}{1 + \sin x \cos x} \right) = -f(x) \)

Now apply the property to the integral \( I \):

\( I = \int_{0}^{\frac{\pi}{2}} f(x) dx = \int_{0}^{\frac{\pi}{2}} f(\frac{\pi}{2} - x) dx \)

\( I = \int_{0}^{\frac{\pi}{2}} -f(x) dx \)

Since the integral of a constant times a function is the constant times the integral of the function:

\( I = - \int_{0}^{\frac{\pi}{2}} f(x) dx \)

We know that \( I = \int_{0}^{\frac{\pi}{2}} f(x) dx \), so:

\( I = -I \)

Adding \( I \) to both sides gives:

\( I + I = 0 \)

\( 2I = 0 \)

Dividing by 2:

\( I = 0 \)

Thus, the value of the definite integral is 0.

Revision Table: Key Concepts

Additional Information on Definite Integral Properties

Definite integral properties are very useful for evaluating integrals, especially when direct integration is difficult. The property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \) is particularly powerful. Let's explore why it works.

Consider the substitution \( u = a - x \). Then \( du = -dx \). When \( x = 0 \), \( u = a \). When \( x = a \), \( u = a - a = 0 \). So, the integral \( \int_{0}^{a} f(x) dx \) becomes:

\( \int_{a}^{0} f(a-u) (-du) \)

Using the property \( \int_{b}^{a} g(x) dx = - \int_{a}^{b} g(x) dx \):

\( - \int_{0}^{a} f(a-u) (-du) = - (- \int_{0}^{a} f(a-u) du) = \int_{0}^{a} f(a-u) du \)

Since \( u \) is just a dummy variable of integration, we can replace it with \( x \):

\( \int_{0}^{a} f(a-u) du = \int_{0}^{a} f(a-x) dx \)

This proves the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \).

In our problem, we applied this property with \( a = \frac{\pi}{2} \). The integrand \( f(x) = \frac{\sin x - \cos x}{1 + \sin x \cos x} \) turned out to be such that \( f(\frac{\pi}{2} - x) = -f(x) \). When this condition \( f(a-x) = -f(x) \) holds for an integral from \( 0 \) to \( a \), the integral evaluates to 0. This is because \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx = \int_{0}^{a} -f(x) dx = - \int_{0}^{a} f(x) dx \), leading to \( 2 \int_{0}^{a} f(x) dx = 0 \), and thus the integral is 0.