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The value of $\int \cos (\log x) d x$ is
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The correct answer is:
$\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C$
Let $I=\int \cos (\log x) \cdot 1 d x...(i)$
Use integral by parts, $\begin{aligned} I=& \cos (\log x) \cdot x-\int[-\sin (\log x)] \cdot \frac{1}{x} \cdot x d x \\=& x \cdot \cos (\log x)+\int \sin (\log x) \cdot 1 d x \\=& x \cdot \cos (\log x)+[\sin (\log x) \cdot x\\ \left.\quad-\int \cos (\log x) \cdot \frac{1}{x} \cdot x d x\right]+C \\=& x \cdot \cos (\log x)+[x \sin (\log x)\\ \left.-\int \cos (\log x) d x\right]+C \\=& x\{\sin (\log x)+\cos (\log x)\}-I+C \\ \quad[\text { from Eq. (i) }] \\ \Rightarrow \quad & I=\frac{x}{2}\{\sin (\log x)+\cos (\log x)\}+C \end{aligned}$
Use integral by parts, $\begin{aligned} I=& \cos (\log x) \cdot x-\int[-\sin (\log x)] \cdot \frac{1}{x} \cdot x d x \\=& x \cdot \cos (\log x)+\int \sin (\log x) \cdot 1 d x \\=& x \cdot \cos (\log x)+[\sin (\log x) \cdot x\\ \left.\quad-\int \cos (\log x) \cdot \frac{1}{x} \cdot x d x\right]+C \\=& x \cdot \cos (\log x)+[x \sin (\log x)\\ \left.-\int \cos (\log x) d x\right]+C \\=& x\{\sin (\log x)+\cos (\log x)\}-I+C \\ \quad[\text { from Eq. (i) }] \\ \Rightarrow \quad & I=\frac{x}{2}\{\sin (\log x)+\cos (\log x)\}+C \end{aligned}$
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