MHT CET · Maths · Indefinite Integration
\(\int[\sin |\log x|+\cos |\log x|] d x=\)
- A \(\sin |\log x|+c\)
- B \(\cos |\log x|+c\)
- C \(x \cos |\log x|+c\)
- D \(x \sin |\log x|+c\)
Answer & Solution
Correct Answer
(D) \(x \sin |\log x|+c\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{I}=\int[\sin |\log \mathrm{x}|+\cos |\log \mathrm{x}|] \mathrm{dx}\)
Put \(\log \mathrm{x}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt}\)
\(\therefore \mathrm{I}=\int(\sin +\cos t) \mathrm{e}^{\mathrm{t}} d t\)
\(=\int \mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}+\cos t) d t=\mathrm{e}^{\mathrm{t}} \sin \mathrm{t}+\mathrm{c}\) \(=\mathrm{x} \sin |\log \mathrm{x}|+\mathrm{c}\)
Put \(\log \mathrm{x}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt}\)
\(\therefore \mathrm{I}=\int(\sin +\cos t) \mathrm{e}^{\mathrm{t}} d t\)
\(=\int \mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}+\cos t) d t=\mathrm{e}^{\mathrm{t}} \sin \mathrm{t}+\mathrm{c}\) \(=\mathrm{x} \sin |\log \mathrm{x}|+\mathrm{c}\)
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