MHT CET · Maths · Indefinite Integration
The value of \(\int \cos (\log x) d x\) is
- A \(\frac{1}{2}[\sin (\log x)+\cos (\log x)]+C\)
- B \(\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C\)
- C \(\frac{x}{2}[\sin (\log x)-\cos (\log x)]+C\)
- D \(\frac{1}{2}[\sin (\log x)-\cos (\log x)]+C\)
Answer & Solution
Correct Answer
(B) \(\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \cos (\log x) \cdot 1 d x...(i)\)
Use integral by parts,
\(I=\cos (\log x) \cdot x-\int[-\sin (\log x)] \cdot \frac{1}{x} \cdot x d x \)
\(=x \cdot \cos (\log x)+\int \sin (\log x) \cdot 1 d x \)
\(=x \cdot \cos (\log x)+[\sin (\log x) \cdot x\) \( \left.-\int \cos (\log x) \cdot \frac{1}{x} \cdot x d x\right]+C \)
\(=x \cdot \cos (\log x)+[x \sin (\log x)\) \( \left.-\int \cos (\log x) d x\right]+C \)
\(=x\{\sin (\log x)+\cos (\log x)\}-I+C \) \(\quad[\text { from Eq. (i) }] \)
\( \Rightarrow I=\frac{x}{2}\{\sin (\log x)+\cos (\log x)\}+C\)
Use integral by parts,
\(I=\cos (\log x) \cdot x-\int[-\sin (\log x)] \cdot \frac{1}{x} \cdot x d x \)
\(=x \cdot \cos (\log x)+\int \sin (\log x) \cdot 1 d x \)
\(=x \cdot \cos (\log x)+[\sin (\log x) \cdot x\) \( \left.-\int \cos (\log x) \cdot \frac{1}{x} \cdot x d x\right]+C \)
\(=x \cdot \cos (\log x)+[x \sin (\log x)\) \( \left.-\int \cos (\log x) d x\right]+C \)
\(=x\{\sin (\log x)+\cos (\log x)\}-I+C \) \(\quad[\text { from Eq. (i) }] \)
\( \Rightarrow I=\frac{x}{2}\{\sin (\log x)+\cos (\log x)\}+C\)
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