The value of integral \((\pi / 4) \int_{0}[2 /(\sec x+\operatorname{cosec} x+\tan x+\cot x)] d x\) is \(\ldots \ldots\) (a) 0 (b) \(1-(\pi / 4)\) (c) \((\pi / 4)+1\) (d) \((\pi / 2)+1\)

First, let's simplify the integrand by expressing \(\sec x\), \(\csc x\), \(\tan x\), and \(\cot x\) in terms of sine and cosine functions: \[\frac{1}{\cos x}, \frac{1}{\sin x}, \frac{\sin x}{\cos x}, \frac{\cos x}{\sin x}\] The integrand then becomes: \[\frac{2}{\frac{1}{\cos x} + \frac{1}{\sin x} + \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}}\] Now, we can find a common denominator and rewrite the expression: \[\frac{2}{\frac{\sin x \cos x + \cos x \sin x + (\sin^2 x + \cos^2 x)}{\sin x\cos x}}\] Simplify the expression: \[\frac{2\sin x\cos x}{2\sin x\cos x + (\sin^2 x + \cos^2 x)}\] Since \(\sin^2 x + \cos^2 x = 1\), the integrand simplifies to: \[\frac{2\sin x\cos x}{2\sin x\cos x + 1}\]

Let \(u = \sin x\cos x\), then \(du = (\cos^2 x - \sin^2 x)dx\). We will now substitute the variable to make the integral more manageable: \[\frac{\pi}{4} \int \frac{2u}{2u + 1} du\]

Now that we have successfully simplified the integral, we can proceed to find the antiderivative of the given integrand: \[\frac{\pi}{4} \int \frac{2u}{2u + 1} du\] First, we will use the substitution method again. Let \(v = 2u + 1\), then \(dv = 2du\). So, the integral becomes: \[\frac{\pi}{4} \int \frac{v}{v} \frac{1}{2} dv\] Integrate with respect to \(v\): \[\frac{\pi}{4} \frac{1}{2} \int dv = \frac{\pi}{8}v + C\] Now, we'll reverse the substitution: \[\frac{\pi}{8}(2u + 1) + C = \frac{\pi}{8}(2\sin x\cos x + 1) + C\]

Now, let's apply the limits of integration from \(0\) to \(\frac{\pi}{2}\): \[\frac{\pi}{8}\Big[\frac{(2\sin x\cos x + 1)}{(2\sin x\cos x)}\Big]\Bigg|_0^{\frac{\pi}{2}} = \frac{\pi}{8}\Bigg(\frac{2\sin \frac{\pi}{2}\cos \frac{\pi}{2} + 1}{2\sin \frac{\pi}{2}\cos \frac{\pi}{2}} - \frac{2\sin0\cos0+1}{2\sin0\cos0}\Bigg)\] Since \(\sin \frac{\pi}{2} = 1\) and \(\cos \frac{\pi}{2} = 0\), and also \(\sin 0 = 0\) and \(\cos 0 = 1\), the expression simplifies to: \[\frac{\pi}{8}\Bigg(\frac{1}{0} - \frac{2\cdot0\cdot1+1}{2\cdot0\cdot1}\Bigg)\] The first term is \(\frac{1}{0}\), which is undefined. However, since we know that \(\lim_{x\to\frac{\pi}{2}} \frac{2\sin x\cos x + 1}{2\sin x\cos x}\) is well-behaved, the second term is just zero, so the integral expression evaluates to: \[\frac{\pi}{8} \cdot \lim_{x\to\frac{\pi}{2}} \frac{1}{0} - 0 = 0\] Thus, the value of the integral is equal to 0. The answer is: (a) 0