Q93 / 150 · MHT CET 2025 (21 Apr Shift 2) · Mathematics

$\int \cos \left(\frac{x}{16}\right) \cdot \cos \left(\frac{x}{8}\right) \cdot \cos \left(\frac{x}{4}\right) \cdot \sin \left(\frac{x}{16}\right) \mathrm{dx}=$

  • A
    $\frac{\cos 16 x}{256}+c \quad$, where $c$ is the constant of integration
  • B
    $\frac{-\cos 16 x}{256}+c$, where $c$ is the constant of integration
  • C
    $\frac{\sin 16 x}{256}+c, \quad$ where $c$ is the constant of integration
  • Correct answer
    D
    $\frac{-\cos \left(\frac{x}{2}\right)}{4}+c$, where $c$ is the constant of integration

Solution

The integral is evaluated by applying the identity $\sin\theta \cos\theta = \frac{1}{2}\sin(2\theta)$ repeatedly.

Combining $\sin\left(\frac{x}{16}\right)\cos\left(\frac{x}{16}\right)$ yields $\frac{1}{2}\sin\left(\frac{x}{8}\right)$.

Combining $\sin\left(\frac{x}{8}\right)\cos\left(\frac{x}{8}\right)$ yields $\frac{1}{2}\sin\left(\frac{x}{4}\right)$.

The remaining product becomes $\frac{1}{4} \int \sin\left(\frac{x}{4}\right)\cos\left(\frac{x}{4}\right) \, \mathrm{dx}$, and combining these gives $\frac{1}{8} \int \sin\left(\frac{x}{2}\right) \, \mathrm{dx}$.

Integration yields $-\frac{1}{4} \cos\left(\frac{x}{2}\right) + c$, matching option D.

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

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