Question:
If \(\int_0^{\pi/2} \frac{\cot x}{\cot x + \operatorname{cosec} x}\,dx = m(\pi + n)\), then \(m \cdot n\) is equal to
If \(\int_0^{\pi/2} \frac{\cot x}{\cot x + \operatorname{cosec} x}\,dx = m(\pi + n)\), then \(m \cdot n\) is equal to
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Write \(\frac{\cot x}{\cot x+\operatorname{cosec} x} = \frac{\cos x}{1+\cos x}\), then use \(1+\cos x = 2\cos^2(x/2)\).
Updated On: Nov 5, 2026
- \(-1\)
- \(1\)
- \(-\frac{1}{2}\)
- \(\frac{1}{2}\)
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The Correct Option is A
Solution and Explanation
\(\frac{\cot x}{\cot x+\operatorname{cosec} x} = \frac{\cos x/\sin x}{(\cos x+1)/\sin x} = \frac{\cos x}{1+\cos x} = 1 - \frac{1}{1+\cos x}\).
So \(I = \int_0^{\pi/2} 1\,dx - \int_0^{\pi/2}\frac{dx}{2\cos^2(x/2)} = \frac{\pi}{2} - \left[\tan\frac{x}{2}\right]_0^{\pi/2} = \frac{\pi}{2} - 1\).
\(\frac{\pi}{2}-1 = \frac{1}{2}(\pi - 2)\), so \(m=\frac{1}{2},\ n=-2\).
\(m\cdot n = \frac{1}{2}\times(-2) = -1\).
Answer: (1) \(-1\)
So \(I = \int_0^{\pi/2} 1\,dx - \int_0^{\pi/2}\frac{dx}{2\cos^2(x/2)} = \frac{\pi}{2} - \left[\tan\frac{x}{2}\right]_0^{\pi/2} = \frac{\pi}{2} - 1\).
\(\frac{\pi}{2}-1 = \frac{1}{2}(\pi - 2)\), so \(m=\frac{1}{2},\ n=-2\).
\(m\cdot n = \frac{1}{2}\times(-2) = -1\).
Answer: (1) \(-1\)
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