The value of _-1/√2^1/√2[((x+1)/(x-1))^2+((x-1)/(x+1))^2-2]^1/2dx is

Question

The value of $\displaystyle\int_{-1/\sqrt2}^{1/\sqrt2}\left[\left(\frac{x+1}{x-1}\right)^2+\left(\frac{x-1}{x+1}\right)^2-2\right]^{1/2}dx$ is

Accepted Answer

$$\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} $$

$$ = \sqrt {{{\left( {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right)}^2}} $$

$$ = \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|$$

$$ = \left| {{{{{(x + 1)}^2} - {{(x - 1)}^2}} \over {{x^2} - 1}}} \right|$$

$$ = \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|$$

$$ = \left| {{{4x} \over {{x^2} - 1}}} \right|$$

$$\therefore$$ $$I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx} $$

$$I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $$

Let $$f(x) = \left| {{{4x} \over {{x^2} - 1}}} \right|$$

$$\therefore$$ $$f( - x) = \left| {{{4( - x)} \over {{{( - x)}^2} - 1}}} \right| = \left| {{{ - 4x} \over {{x^2} - 1}}} \right| = \left| {{{4x} \over {{x^2} - 1}}} \right|$$

$$ \Rightarrow f(x) = f( - x)$$

$$\therefore$$ f(x) is a even function.

$$\therefore$$ $$I = 2\int_0^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $$

Using property, If f(x) is an even function then,

$$\int_{ - a}^a {f(x) = 2\int_0^a {f(x)dx} } $$

x > 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\Rightarrow$$ x2 > 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\Rightarrow$$ x2 - 1

Solutions · 1