Q89 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics

$\int \left(e^{\log(\sin x)} + \cos x\right) x\, dx = $

  • A
    $x(\sin x + \cos x) + (\sin x - \cos x) + c$
  • B
    $x(\sin x - \cos x) + (\sin x - \cos x) + c$
  • C
    $x(\sin x + \cos x) + (\sin x + \cos x) + c$
  • Correct answer
    D
    $x(\sin x - \cos x) + (\sin x + \cos x) + c$

Solution

Using the property $e^{\log f(x)} = f(x)$, the given integral simplifies to:

$\int x(\sin x + \cos x) dx$

Applying integration by parts, taking $x$ as the first function and $(\sin x + \cos x)$ as the second function:

$\int x(\sin x + \cos x) dx = x \int (\sin x + \cos x) dx - \int \left( \dfrac{d}{dx}(x) \int (\sin x + \cos x) dx \right) dx$

$= x(\sin x - \cos x) - \int 1 \cdot (\sin x - \cos x) dx$

$= x(\sin x - \cos x) - (-\cos x - \sin x) + c$

$= x(\sin x - \cos x) + (\sin x + \cos x) + c$

Answer: $x(\sin x - \cos x) + (\sin x + \cos x) + c$

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

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