A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) 0
Correct Answer: C
Solution :
| Let \[I=\int_{0}^{\infty }{\frac{1}{\left( 1+{{x}^{2020}} \right)\left( 1+{{x}^{2}} \right)}dx}\] Put \[x=\frac{1}{t}dx=\frac{-1}{{{t}^{2}}}dt\] |
| \[\therefore I=\int_{0}^{\infty }{\frac{1}{\left( 1+\frac{1}{{{t}^{2020}}} \right)\left( 1+\frac{1}{{{t}^{2}}} \right)}-\frac{dt}{{{t}^{2}}}}\]\[\Rightarrow \,\,I=\int_{0}^{\infty }{\frac{{{t}^{2020}}dt}{\left( {{t}^{2020}}+1 \right)\left( {{t}^{2}}+1 \right)}}\]\[\Rightarrow \,I=\int_{0}^{\infty }{\frac{{{x}^{2020}}dx}{\left( 1+{{x}^{2020}} \right)\left( 1+{{x}^{2}} \right)}}\]\[\Rightarrow \,2I=\int_{0}^{\infty }{\frac{{{x}^{2020}}+1}{\left( 1+{{x}^{2020}} \right)\left( 1+{{x}^{2}} \right)}}dx\]\[\Rightarrow \,2I=\int_{0}^{\infty }{\frac{1}{\left( 1+{{x}^{2}} \right)}}dx\Rightarrow 2I=\left[ {{\tan }^{-1}}x \right]_{0}^{\infty }\]\[\Rightarrow \,2I=\left( {{\tan }^{-1}}\infty -{{\tan }^{-1}}0 \right)\]\[\Rightarrow \,2I=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}\] |
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