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${x^2}f(x) + f\left( {\frac{1}{x}} \right) = 0,x \ne 0$

$I = \int\limits_{\tan \alpha }^{\cot \alpha } {f(x)dx} = ?$

We have, $f(x) =  - \frac{1}{{{x^2}}}f\left( {\frac{1}{x}} \right)$

$\therefore I = \int\limits_{\tan \alpha }^{\cot \alpha } { - \frac{1}{{{x^2}}}f\left( {\frac{1}{x}} \right)dx} $

Let, $\frac{1}{x} = t$

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

$\therefore I = \int\limits_{\cot \alpha }^{\tan \alpha } {f(t)dt} =-I $

$ \Rightarrow I = 0$