Q51 / 125 · BITSAT 2021 · Mathematics

The value of $\int \cos (\log x) \mathrm{d} \mathrm{x}$ is :

  • A
    $\frac{1}{2}[\sin (\log x)+\cos (\log x)]+C$
  • Correct answer
    B
    $\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C$
  • C
    $\frac{x}{2}[\sin (\log x)-\cos (\log x)]+C$
  • D
    $\frac{1}{2}[\sin (\log x)-\cos (\log x)]+C$

Solution

Let \(\mathrm{I}=\int \cos (\log \mathrm{x}) \cdot 1 \mathrm{~d} \mathrm{x} \ldots . . \text { (i) }\) Use integral by parts, \(\begin{aligned} &I=\cos (\log x) \cdot x-\int[-\sin (\log x)] \cdot \frac{1}{x} \cdot x d x \\ &=x \cdot \cos (\log x)+\int \sin (\log x) \cdot 1 d x \\ &=x \cdot \cos (\log x)+\left[\sin (\log x) \cdot x-\int \cos (\log x) \cdot \frac{1}{x} \cdot x d x\right]+C \\ &=x \cdot(\log x)+\left[x \sin (\log x)-\int \cos (\log x) d x\right]+C \\ &=x\{\sin (\log x)+\cos (\log x)\}-I+C[f r o m \text { Eq. (i) }] \\ &\Rightarrow I=\frac{x}{2}\{\sin (\log x)+\cos (\log x)\}+C \end{aligned}\)

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

Relevant Questions