Question

If f(x) is a function satisfying f(1/x) + x^2f(x) = 0, for all non zero x, I find the value of

$\int_{sint}^{cosect} \: f(x) \: dx$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:f\bigg(\dfrac{1}{x} \bigg) + {x}^{2}f(x) \: = \: 0 \:$

can be rewritten as

$\rm :\longmapsto\: {x}^{2}f(x) \: = \: - \: f\bigg(\dfrac{1}{x} \bigg) \:$

$\purple{\rm :\longmapsto\:f(x) = - \: \dfrac{1}{ {x}^{2} }f\bigg(\dfrac{1}{x} \bigg) - - - (1)}$

Now, Given integral is

$\rm :\longmapsto\:\displaystyle\int_{sint}^{cosect} \: f(x) \: dx$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle\int_{sint}^{cosect} \: f(x) \: dx$

On substituting the value of f(x), from equation (1), we get

$\rm :\longmapsto\:I = \:\displaystyle\int_{sint}^{cosect} \: - \: \dfrac{1}{ {x}^{2}} \: f\bigg(\dfrac{1}{x} \bigg) \: dx$

To evaluate this integral, we use method of Substitution.

So, we substitute

$\red{\rm :\longmapsto\:\dfrac{1}{x} = y \: }$

$\red{\rm :\longmapsto\: - \: \dfrac{1}{ {x}^{2} } \: dx = dy \: }$

When we substitute, we have to change the limits too.

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = \dfrac{1}{x} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sint & \sf cosect \\ \\ \sf cosect & \sf sint \end{array}} \\ \end{gathered}$

So, above integral can be reduced to

$\rm :\longmapsto\:I = \:\displaystyle\int_{cosect}^{sint} \: - \: f(y)\: dy$

$\rm :\longmapsto\:I = \: - \:\displaystyle\int_{cosect}^{sint} \: \: f(y)\: dy$

We know, Limit interchanging property

$\boxed{ \tt{ \: \displaystyle\int_{a}^{b}f(x)dx \: = \: - \: \displaystyle\int_{b}^{a}f(x) \: dx \: }}$

So, using this, we get

$\rm :\longmapsto\:I \: = \: - \: \displaystyle\int_{sint}^{cosect} \: f(y) \: dy$

We know, Variable changing property

$\boxed{ \tt{ \: \displaystyle\int_{a}^{b}f(x) \: dx \: = \: \displaystyle\int_{a}^{b}f(y) \: dy \: }}$

$\rm :\longmapsto\:I \: = \: - \: \displaystyle\int_{sint}^{cosect} \: f(x) \: dx$

$\rm :\longmapsto\:I \: = \: - \: I$

$\rm :\longmapsto\:2I \: = \: 0$

$\bf\implies \:I = 0$

Hence,

$\red{\rm \implies\:\boxed{ \tt{ \: \:\displaystyle\int_{sint}^{cosect} \: f(x) \: dx = 0 \: }}}$

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