\(\int_{0}^{\frac{\pi}{2}}\displaystyle\frac{1-\cot x}{\mathrm{cosec}\,x+\cos x}\,dx =\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Multiplying the numerator and the denominator by sin x:
\( \displaystyle \frac{1-\cot x}{\mathrm{cosec}\,x+\cos x} = \frac{\sin x - \cos x}{1 + \sin x\cos x} \)
Let
\( \displaystyle I = \int_{0}^{\pi/2}\frac{\sin x - \cos x}{1 + \sin x\cos x}\,dx \qquad (1) \)
By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\), replace x by \(\frac{\pi}{2} - x\). Then sin x and cos x change places:
\( \displaystyle I = \int_{0}^{\pi/2}\frac{\cos x - \sin x}{1 + \cos x\sin x}\,dx \qquad (2) \)
The integrand in (2) is the negative of the integrand in (1), so
\(I = -I\)
\(2I = 0\)
\(I = 0\)