Question

The value of ${\int }_{1/2}^2\frac{1}{x}{\csc }^{101}(x-\frac{1}{x})dx$ is?

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

Answer 1

The correct option is C $0$

Let $I={\int }_{1/2}^2\frac{1}{x}{\csc }^{101}(x-\frac{1}{x})dx$

Substitute $\frac{1}{x}=t\Rightarrow -\frac{1}{x^2}dx=dt$

$\therefore I=-{\int }_{1/2}^2\frac{1}{t}{\csc }^{101}(t-\frac{1}{t})dt\Rightarrow I=-I\Rightarrow 2I=0\Rightarrow I=0$