Question

${\int }_{1/2}^2\frac{1}{x}cose{c}^{101}(x-\frac{1}{x})dx$ is equal to.

• A. $\frac{1}{4}$
• B. $1$
• C. $0$
• D. $\frac{101}{2}$

Answer 1

The correct option is C $0$

Let $I={\int }_{1/2}^2\frac{1}{x}cose{c}^{101}(x-\frac{1}{x})dx$
Put $\frac{1}{x}=t\Rightarrow -\frac{1}{x^2}dx=dt$
$\therefore I=-{\int }_{1/2}^2\frac{1}{t}cose{c}^{101}(t-\frac{1}{t})dt$
$\Rightarrow I=-I\Rightarrow 2I=0$
$\Rightarrow I=0$