Q83 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
$\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{\sin x - \cos x}{1 + \sin x\cos x}\,dx =$
- Correct answerA$0$
- B$\dfrac{\pi}{12}$
- C$\dfrac{\pi}{24}$
- D$\dfrac{\pi^2}{2}$
Solution
Let $I = \int_{\pi/6}^{\pi/3} \dfrac{\sin x - \cos x}{1 + \sin x\cos x}\,dx$
Using the definite integral property $\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx$, we substitute $x$ with $\dfrac{\pi}{6} + \dfrac{\pi}{3} - x = \dfrac{\pi}{2} - x$.
$I = \int_{\pi/6}^{\pi/3} \dfrac{\sin(\dfrac{\pi}{2} - x) - \cos(\dfrac{\pi}{2} - x)}{1 + \sin(\dfrac{\pi}{2} - x)\cos(\dfrac{\pi}{2} - x)}\,dx$
$I = \int_{\pi/6}^{\pi/3} \dfrac{\cos x - \sin x}{1 + \cos x\sin x}\,dx$
$I = -\int_{\pi/6}^{\pi/3} \dfrac{\sin x - \cos x}{1 + \sin x\cos x}\,dx$
$I = -I$
$2I = 0$
$I = 0$
Answer: $0$
Using the definite integral property $\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx$, we substitute $x$ with $\dfrac{\pi}{6} + \dfrac{\pi}{3} - x = \dfrac{\pi}{2} - x$.
$I = \int_{\pi/6}^{\pi/3} \dfrac{\sin(\dfrac{\pi}{2} - x) - \cos(\dfrac{\pi}{2} - x)}{1 + \sin(\dfrac{\pi}{2} - x)\cos(\dfrac{\pi}{2} - x)}\,dx$
$I = \int_{\pi/6}^{\pi/3} \dfrac{\cos x - \sin x}{1 + \cos x\sin x}\,dx$
$I = -\int_{\pi/6}^{\pi/3} \dfrac{\sin x - \cos x}{1 + \sin x\cos x}\,dx$
$I = -I$
$2I = 0$
$I = 0$
Answer: $0$
Hence, the correct answer is option: 'A'.