Q115 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

$\int_{\frac{1}{3}}^3 \frac{1}{x} \sin \left(\frac{1}{x}-x\right) d x=$

  • Correct answer
    A
    0
  • B
    $\frac{4}{3}$
  • C
    $\frac{5}{3}$
  • D
    $\frac{8}{3}$

Solution

Let $$ \begin{aligned} & I=\int_{1 / 3}^3 \frac{1}{x} \sin \left(\frac{1}{x}-x\right) d x \\ & \text { Put } x=\frac{1}{t} \Rightarrow d x=\frac{-1}{t^2} d t \\ & x=\frac{1}{3}, t=3 \text { and } x=3, t=\frac{1}{3} \\ & \therefore I=\int_3^{1 / 3} t \sin \left(t-\frac{1}{t}\right)\left(\frac{-1}{t^2}\right) d t \\ & I=-\int_3^{1 / 3} \frac{1}{t} \sin \left(t-\frac{1}{t}\right) d t \\ & \Rightarrow \quad I=\int_3^{1 / 3} \frac{1}{t} \sin \left(\frac{1}{t}-t\right) d t \\ & I=-\int_{1 / 3}^3 \frac{1}{t} \sin \left(\frac{1}{t}-t\right) d t \\ & \Rightarrow \quad I=-I \\ & \Rightarrow \quad I=0 \\ & \end{aligned} $$

Hence, the correct answer is option: 'A'.

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