MHT CET · Maths · Indefinite Integration
\(\int\left(\frac{\tan \left(\frac{1}{x}\right)}{x}\right)^2 \mathrm{~d} x=\)
- A \(x-\tan x+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration
- B \(\frac{1}{x}-\tan \left(\frac{1}{x}\right)+c\), where \(\mathrm{c}\) is a constant of integration.
- C \(\frac{1}{x}+\tan \left(\frac{1}{x}\right)+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- D \(x+\tan x+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Answer & Solution
Correct Answer
(B) \(\frac{1}{x}-\tan \left(\frac{1}{x}\right)+c\), where \(\mathrm{c}\) is a constant of integration.
Step-by-step Solution
Detailed explanation
Let \(\mathrm{I}=\int\left(\frac{\tan \left(\frac{1}{x}\right)}{x}\right)^2 \mathrm{~d} x\)
Let \(\frac{1}{x}=\mathrm{t} \Rightarrow \frac{1}{x^2} \mathrm{~d} x=-\mathrm{dt}\)
\(\therefore \mathrm{I} =-\int \tan ^2 \mathrm{t} d \mathrm{t} \)
\( =\int\left(1-\sec ^2 \mathrm{t}\right) \mathrm{dt} \)
\( =\mathrm{t}-\tan \mathrm{t}+\mathrm{c} \)
\( =\frac{1}{x}-\tan \left(\frac{1}{x}\right)+\mathrm{c}\)
Let \(\frac{1}{x}=\mathrm{t} \Rightarrow \frac{1}{x^2} \mathrm{~d} x=-\mathrm{dt}\)
\(\therefore \mathrm{I} =-\int \tan ^2 \mathrm{t} d \mathrm{t} \)
\( =\int\left(1-\sec ^2 \mathrm{t}\right) \mathrm{dt} \)
\( =\mathrm{t}-\tan \mathrm{t}+\mathrm{c} \)
\( =\frac{1}{x}-\tan \left(\frac{1}{x}\right)+\mathrm{c}\)
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