Q90 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics

$\int\sin(\log x)\,dx = $

  • Correct answer
    A
    $\dfrac{x}{2}[\sin(\log x) - \cos(\log x)] + c$
  • B
    $\dfrac{x}{2}[\sin(\log x) + \cos(\log x)] + c$
  • C
    $\dfrac{x}{2}[\cos(\log x) - \sin(\log x)] + c$
  • D
    $\dfrac{x}{4}[\cos(\log x) - \sin(\log x)] + c$

Solution

Let $I = \int \sin(\log x) \, dx$

Using integration by parts, taking $1$ as the second function:

$I = \sin(\log x) \int 1 \, dx - \int \left( \dfrac{d}{dx}(\sin(\log x)) \int 1 \, dx \right) dx$

$I = x \sin(\log x) - \int \cos(\log x) \cdot \dfrac{1}{x} \cdot x \, dx$

$I = x \sin(\log x) - \int \cos(\log x) \, dx$

Applying integration by parts again to the second integral:

$I = x \sin(\log x) - \left[ \cos(\log x) \int 1 \, dx - \int \left( \dfrac{d}{dx}(\cos(\log x)) \int 1 \, dx \right) dx \right]$

$I = x \sin(\log x) - \left[ x \cos(\log x) - \int \left( -\sin(\log x) \cdot \dfrac{1}{x} \cdot x \right) dx \right]$

$I = x \sin(\log x) - x \cos(\log x) - \int \sin(\log x) \, dx$

$I = x \sin(\log x) - x \cos(\log x) - I$

$2I = x [\sin(\log x) - \cos(\log x)]$

$I = \dfrac{x}{2} [\sin(\log x) - \cos(\log x)] + c$

Answer: $\dfrac{x}{2}[\sin(\log x) - \cos(\log x)] + c$

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

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