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Calculus Level 5

∫ 0 1 ( cot ⁡ − 1 ( x 2 + 2 ) ( x 2 + 1 ) ( x 2 + 2 ) ) d x \large \int _{ 0 }^{ 1 }{ \left( \frac { { \cot }^{ -1 }\left( \sqrt { { x }^{ 2 }+2 } \right) }{ \left( { x }^{ 2 }+1 \right) \left( \sqrt { { x }^{ 2 }+2 } \right) } \right) }dx

If the integral above equals to π A B \dfrac { { \pi }^{ A } }{ B } for positive integers A A and B B , find A + B A+B .

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The answer is 34.

Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Aditya Kumar
Sep 20, 2015

I = ∫ 0 1 ( c o t − 1 ( x 2 + 2 ) ( x 2 + 1 ) ( x 2 + 2 ) ) = ∫ 0 1 ( t a n − 1 ( 1 x 2 + 2 ) ( x 2 + 1 ) ( x 2 + 2 ) ) we know that, 1 a t a n − 1 ( 1 a ) = ∫ 0 1 1 y 2 + a 2 d y I = ∫ 0 1 ∫ 0 1 1 ( x 2 + 1 ) ( y 2 + x 2 + 2 ) d y . d x = ∫ 0 1 ∫ 0 1 1 ( x 2 + 1 ) ( 1 y 2 + 1 − 1 y 2 + x 2 + 2 ) d y . d x = 1 2 ∫ 0 1 ∫ 0 1 1 ( x 2 + 1 ) ( y 2 + 1 ) d y . d x = 1 2 ( ∫ 0 1 1 y 2 + 1 d y ) 2 I = π 2 32 I=\int _{ 0 }^{ 1 }{ \left( \frac { { cot }^{ -1 }\left( \sqrt { { x }^{ 2 }+2 } \right) }{ \left( { x }^{ 2 }+1 \right) \left( \sqrt { { x }^{ 2 }+2 } \right) } \right) } \\ \quad =\int _{ 0 }^{ 1 }{ \left( \frac { { tan }^{ -1 }\left( \frac { 1 }{ \sqrt { { x }^{ 2 }+2 } } \right) }{ \left( { x }^{ 2 }+1 \right) \left( \sqrt { { x }^{ 2 }+2 } \right) } \right) } \\\mbox{ we know that,}\quad \\ \frac { 1 }{ a } { tan }^{ -1 }\left( \frac { 1 }{ a } \right) =\int _{ 0 }^{ 1 }{ \frac { 1 }{ { y }^{ 2 }+{ a }^{ 2 } } dy } \\ I=\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( { x }^{ 2 }+1 \right) \left( { y }^{ 2 }+{ x }^{ 2 }+2 \right) } dy.dx } } \\ \quad =\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( { x }^{ 2 }+1 \right) } \left( \frac { 1 }{ { y }^{ 2 }+1 } -\frac { 1 }{ { y }^{ 2 }+{ x }^{ 2 }+2 } \right) dy.dx } } \\ \quad =\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( { x }^{ 2 }+1 \right) \left( { y }^{ 2 }+1 \right) } dy.dx } } \\ \quad =\frac { 1 }{ 2 } { \left( \int _{ 0 }^{ 1 }{ \frac { 1 }{ { y }^{ 2 }+1 } dy } \right) }^{ 2 }\\ I=\frac { { \pi }^{ 2 } }{ 32 }

12 Helpful 0 Interesting 0 Brilliant 0 Confused

Great work! Brilliant!

Kartik Sharma - 5 years, 7 months ago

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Thanks! What was ur method to solve?

Aditya Kumar - 5 years, 7 months ago

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My method was way longer - with the series and all.

Kartik Sharma - 5 years, 7 months ago

Wow! 　　　　　　　　　　　　　　　　　　　　　　　　　　.

Kenny Lau - 5 years, 8 months ago

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Thanks!

.

Aditya Kumar - 5 years, 8 months ago

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Better if you use \mbox{we know that} :)

Kenny Lau - 5 years, 7 months ago

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@Kenny Lau – What would that do?

Aditya Kumar - 5 years, 7 months ago

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@Aditya Kumar – Make the "we know that" not italics.

Kenny Lau - 5 years, 7 months ago

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@Kenny Lau – Done! 

                                             :)

Aditya Kumar - 5 years, 7 months ago

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@Aditya Kumar – And.... put a backslash before all functions, i.e. "\tan" instead of "tan".

Kenny Lau - 5 years, 7 months ago

@Abhishek Bakshi see this and comment.

Aditya Kumar - 5 years, 8 months ago

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