WBJEE · Maths · Indefinite Integration
\(\int \cos (\log x) d x=F(x)+C,\) where \(C\) is an arbitrary constant. Here, \(F(x)\) is equal to
- A \(x[\cos (\log x)+\sin (\log x)]\)
- B \(x[\cos (\log x)-\sin (\log x)]\)
- C \(\frac{x}{2}[\cos (\log x)+\sin (\log x)]\)
- D \(\frac{x}{2}[\cos (\log x)-\sin (\log x)]\)
Answer & Solution
Correct Answer
(C) \(\frac{x}{2}[\cos (\log x)+\sin (\log x)]\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \cos (\log x) d x\) Put \(\log x=t\) \(\Rightarrow \quad x=e^{t}\) \(\begin{aligned} \therefore \quad d x &=e^{t} d t \\ \therefore \quad I &=\int e^{t} \cos t d t \\ &=\frac{e^{t}}{1^{2}+1^{2}}[\cos t+\sin t]+C \end{aligned}\)…
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