The value of the integral _-1^1_e(√1-x+√1+x)dx is equal to :

Question

The value of the integral $\displaystyle\int_{-1}^1\log_e\left(\sqrt{1-x}+\sqrt{1+x}\right)dx$ is equal to :

Accepted Answer

$$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt 1  - x + \sqrt {1 + x} )dx} $$

We know, $$\int\limits_{ - a}^a {f(x)dx = 2\int\limits_0^a {f(x)dx} } $$

So, $$2\int\limits_0^1 {{{\log }_e}(\sqrt 1  - x + \sqrt {1 + x} )dx} $$

$$l = 2\int\limits_0^1 {{{\log }_e}(\sqrt 1  - x + \sqrt {1 + x} ).1dx} $$

$$ \Rightarrow {l \over 2} = {\log _e}(\sqrt {1 - x}  + \sqrt {1 + x} ).\,x]_0^1 - \int\limits_0^1 {{{{1 \over {2\sqrt {1 + x} }} - {1 \over {2\sqrt {1 - x} }}} \over {\sqrt {1 - x}  + \sqrt {1 + x} }}.\,x\,dx} $$

$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2  - {1 \over 2}\int\limits_0^1 {\left( {{{\sqrt {1 - x}  - \sqrt {1 + x} } \over {\sqrt {1 - x}  + \sqrt {1 + x} }}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$

$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2  - {1 \over 2}\int\limits_0^1 {\left( {{{(1 - x) + (1 + x) - 2\sqrt {1 - {x^2}} } \over {(1 - x) - (1 + x)}}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$

$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2  - {1 \over 2}\int\limits_0^1 {{{2(1 - \sqrt {1 - {x^2}} )} \over { - 2x}}.{x \over {\sqrt {1 - {x^2}} }}dx} $$

$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2  + {1 \over 2}\int\limits_0^1 {\left( {{1 \over {\sqrt {1 - {x^2}} }} - 1} \right)dx} $$

$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2  + {1 \over 2}[{\sin ^{ - 1}}x - x]_0^1$$

$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2  + {1 \over 2}\left( {{\pi  \over 2} - 1} \right)$$

$$\therefore$$ $$l = {\log _e}2 + {\pi  \over 2} - 1$$

Solutions · 1