Q92 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics

If $\int_{\pi/6}^{\pi/3} \dfrac{1}{1+\sin x + \cos x}dx = \log 2$, then the value of $\int_{\pi/6}^{\pi/3} \dfrac{\cos x}{1+\sin x + \cos x}dx = $ ............

  • Correct answer
    A
    $\dfrac{\pi}{12} - \dfrac{1}{2}\log 2$
  • B
    $\dfrac{\pi}{12} - \dfrac{1}{3}\log 2$
  • C
    $\dfrac{\pi}{6} - \dfrac{1}{2}\log 2$
  • D
    $\dfrac{\pi}{12} - \dfrac{1}{4}\log 2$

Solution

Let $I = \int_{\pi/6}^{\pi/3} \dfrac{\cos x}{1+\sin x + \cos x}dx$

Using the definite integral property $\int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx$, we substitute $x$ with $\dfrac{\pi}{6} + \dfrac{\pi}{3} - x = \dfrac{\pi}{2} - x$:

$I = \int_{\pi/6}^{\pi/3} \dfrac{\cos(\pi/2 - x)}{1+\sin(\pi/2 - x) + \cos(\pi/2 - x)}dx$

$I = \int_{\pi/6}^{\pi/3} \dfrac{\sin x}{1+\cos x + \sin x}dx$

Adding the two expressions for $I$:

$2I = \int_{\pi/6}^{\pi/3} \dfrac{\sin x + \cos x}{1+\sin x + \cos x}dx$

$2I = \int_{\pi/6}^{\pi/3} \dfrac{1+\sin x + \cos x - 1}{1+\sin x + \cos x}dx$

$2I = \int_{\pi/6}^{\pi/3} \left(1 - \dfrac{1}{1+\sin x + \cos x}\right)dx$

$2I = \int_{\pi/6}^{\pi/3} 1 dx - \int_{\pi/6}^{\pi/3} \dfrac{1}{1+\sin x + \cos x}dx$

Given that $\int_{\pi/6}^{\pi/3} \dfrac{1}{1+\sin x + \cos x}dx = \log 2$, we substitute this value:

$2I = [x]_{\pi/6}^{\pi/3} - \log 2$

$2I = \left(\dfrac{\pi}{3} - \dfrac{\pi}{6}\right) - \log 2$

$2I = \dfrac{\pi}{6} - \log 2$

$I = \dfrac{\pi}{12} - \dfrac{1}{2}\log 2$

Answer: $\dfrac{\pi}{12} - \dfrac{1}{2}\log 2$

Hence, the correct answer is option: 'A'.

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