JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 21)
If $$2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,$$
then $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$ is equalto :
then $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$ is equalto :
log4
$${\pi \over 2}$$ + log2
log2
$${\pi \over 2}$$ $$-$$ log4
Explanation
Given,
$$2\int\limits_0^1 {{{\tan }^{ - {1_x}\,{d_x}}}} = \int\limits_0^1 {{{\cot }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$
= $$\int\limits_0^1 {\left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} \right)} dx$$
$$ \Rightarrow $$$$\,\,\,$$$$2\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx = \int\limits_0^1 {{\pi \over 2}} dx - \int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$
$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = \int\limits_0^1 {{\pi \over 2}} dx - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,dx} $$
$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = {\pi \over 2} - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,\,dx} $$
Let I = $$\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx$$
= $$\left[ {\left( {{{\tan }^{ - 1}}x} \right)x} \right]_0^1 - \int\limits_0^1 {{1 \over {1 + {x^2}}}} x\,dx$$
= $${\pi \over 4} - {1 \over {20}}\left[ {\log \left| {1 + {x^2}} \right|} \right]_0^1$$
= $${\pi \over 4} - {1 \over 2}\log 2$$
$$\therefore\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)\,dx$$
= $${\pi \over 2} - 2\left( {{\pi \over 4} - {1 \over 2}\log 2} \right)$$
= $${\pi \over 2} - {\pi \over 2} + \log 2$$
= log2
$$2\int\limits_0^1 {{{\tan }^{ - {1_x}\,{d_x}}}} = \int\limits_0^1 {{{\cot }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$
= $$\int\limits_0^1 {\left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} \right)} dx$$
$$ \Rightarrow $$$$\,\,\,$$$$2\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx = \int\limits_0^1 {{\pi \over 2}} dx - \int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$
$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = \int\limits_0^1 {{\pi \over 2}} dx - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,dx} $$
$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = {\pi \over 2} - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,\,dx} $$
Let I = $$\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx$$
= $$\left[ {\left( {{{\tan }^{ - 1}}x} \right)x} \right]_0^1 - \int\limits_0^1 {{1 \over {1 + {x^2}}}} x\,dx$$
= $${\pi \over 4} - {1 \over {20}}\left[ {\log \left| {1 + {x^2}} \right|} \right]_0^1$$
= $${\pi \over 4} - {1 \over 2}\log 2$$
$$\therefore\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)\,dx$$
= $${\pi \over 2} - 2\left( {{\pi \over 4} - {1 \over 2}\log 2} \right)$$
= $${\pi \over 2} - {\pi \over 2} + \log 2$$
= log2
Comments (0)
