JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 3)
The value of the integral $$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx} $$ is equal to:
$${1 \over 2}{\log _e}2 + {\pi \over 4} - {3 \over 2}$$
$$2{\log _e}2 + {\pi \over 4} - 1$$
$${\log _e}2 + {\pi \over 2} - 1$$
$$2{\log _e}2 + {\pi \over 2} - {1 \over 2}$$
Explanation
$$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx} $$
We know, $$\int\limits_{ - a}^a {f(x)dx = 2\int\limits_0^a {f(x)dx} } $$
So, $$2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx} $$
$$l = 2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} ).1dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}(\sqrt {1 - x} + \sqrt {1 + x} ).\,x]_0^1 - \int\limits_0^1 {{{{1 \over {2\sqrt {1 + x} }} - {1 \over {2\sqrt {1 - x} }}} \over {\sqrt {1 - x} + \sqrt {1 + x} }}.\,x\,dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{\sqrt {1 - x} - \sqrt {1 + x} } \over {\sqrt {1 - x} + \sqrt {1 + x} }}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{(1 - x) + (1 + x) - 2\sqrt {1 - {x^2}} } \over {(1 - x) - (1 + x)}}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {{{2(1 - \sqrt {1 - {x^2}} )} \over { - 2x}}.{x \over {\sqrt {1 - {x^2}} }}dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\int\limits_0^1 {\left( {{1 \over {\sqrt {1 - {x^2}} }} - 1} \right)dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}[{\sin ^{ - 1}}x - x]_0^1$$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\left( {{\pi \over 2} - 1} \right)$$
$$\therefore$$ $$l = {\log _e}2 + {\pi \over 2} - 1$$
We know, $$\int\limits_{ - a}^a {f(x)dx = 2\int\limits_0^a {f(x)dx} } $$
So, $$2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx} $$
$$l = 2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} ).1dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}(\sqrt {1 - x} + \sqrt {1 + x} ).\,x]_0^1 - \int\limits_0^1 {{{{1 \over {2\sqrt {1 + x} }} - {1 \over {2\sqrt {1 - x} }}} \over {\sqrt {1 - x} + \sqrt {1 + x} }}.\,x\,dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{\sqrt {1 - x} - \sqrt {1 + x} } \over {\sqrt {1 - x} + \sqrt {1 + x} }}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{(1 - x) + (1 + x) - 2\sqrt {1 - {x^2}} } \over {(1 - x) - (1 + x)}}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {{{2(1 - \sqrt {1 - {x^2}} )} \over { - 2x}}.{x \over {\sqrt {1 - {x^2}} }}dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\int\limits_0^1 {\left( {{1 \over {\sqrt {1 - {x^2}} }} - 1} \right)dx} $$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}[{\sin ^{ - 1}}x - x]_0^1$$
$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\left( {{\pi \over 2} - 1} \right)$$
$$\therefore$$ $$l = {\log _e}2 + {\pi \over 2} - 1$$
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