Q76 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics

$\int_0^{\pi/2} \dfrac{3\sin x + 4\cos x}{\sin x + \cos x}\, dx =$

  • A
    $\dfrac{7\pi}{2}$
  • B
    $\dfrac{\pi}{4}$
  • Correct answer
    C
    $\dfrac{7\pi}{4}$
  • D
    $7\pi$

Solution

Let $I = \int_0^{\pi/2} \dfrac{3\sin x + 4\cos x}{\sin x + \cos x}\, dx$

Using the definite integral property $\int_0^a f(x)\, dx = \int_0^a f(a-x)\, dx$, we get:

$I = \int_0^{\pi/2} \dfrac{3\sin(\pi/2 - x) + 4\cos(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)}\, dx$

$I = \int_0^{\pi/2} \dfrac{3\cos x + 4\sin x}{\cos x + \sin x}\, dx$

Adding the two equations for $I$:

$2I = \int_0^{\pi/2} \dfrac{3\sin x + 4\cos x + 3\cos x + 4\sin x}{\sin x + \cos x}\, dx$

$2I = \int_0^{\pi/2} \dfrac{7(\sin x + \cos x)}{\sin x + \cos x}\, dx$

$2I = \int_0^{\pi/2} 7\, dx$

$2I = 7[x]_0^{\pi/2} = \dfrac{7\pi}{2}$

$I = \dfrac{7\pi}{4}$

Answer: $\dfrac{7\pi}{4}$

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

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