\[ \large \int _{ -1 }^{ 1 }{ \left( \cot^{-1}{ \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } } \right) } \left( \cot^{-1}{ \frac { x }{ \sqrt { 1-{ \left( { x }^{ 2 } \right) }^{ \left| x \right| } } } } \right) dx\quad =\quad \frac { { \pi }^{ 2 }\left( \sqrt { a } -\sqrt { b } \right) }{ \sqrt { c } } \]

Please help me. I've been working on this integral since last week but to no avail.

With \(a,b,c\) are positive integers in their lowest form, find \(a+b+c\).

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## Comments

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TopNewest@Abhimanyu Swami Since it's really late, I'll just write part of the solution and will complete it sometime tomorrow. \[\] Use the property \[\int_a^b f(x)dx = \int_a^b f(a+b-x)dx\] Notice that we can thus write \[\] \[I=\int _{ -1 }^{ 1 }{ \left( \cot^{-1}{ \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } } \right) } \left( \cot^{-1}{ \frac { x }{ \sqrt { 1-{ \left( { x }^{ 2 } \right) }^{ \left| x \right| } } } } \right) dx\quad\] \[\] \[I=\int _{ -1 }^{ 1 }{ \left( \cot^{-1}{ \frac { 1 }{ \sqrt { 1-{ (-x) }^{ 2 } } } } \right) } \left( \cot^{-1}{ \frac { (-x) }{ \sqrt { 1-{ \left( { (-x) }^{ 2 } \right) }^{ \left| -x \right| } } } } \right) dx\quad\] Add both to get \[\Longrightarrow I=\dfrac{1}{2}\int_{-1}^1 \dfrac{\pi^2}{2} - \pi \tan^{-1}\left(\dfrac{1}{\sqrt{1-x^2}}\right)dx\] \[\] The rest should then become easy. \[\] Your final result seems to be \[I = \dfrac{\pi^2}{2} \left(\sqrt{2}-1\right)\] \[\] \[\]Hope this helps.

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YEAH THANKS FOR THAT.

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Welcome! Glad to have been of help! On a side note, you shouldn't write everything in all caps since it's considered rude on the Net. I of course have no objection, but others might...

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Woah! Nice! I didn't know of that property.

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Thanks very much!

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In fact, this property is very useful when dealing with Inverse Trig Integrals, and is also a useful thing to check for when you see an integral and nothing strikes you about how to solve it (this experience of nothing striking you must be a rarity for you!).

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@Kartik Sharma Why don't you post some problems on Integration? We could all learn an enormous amount from you. By the way, try this-you'll hopefully like it:)

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@Ishan Dasgupta Samarendra

which coaching are u referring to!?Log in to reply

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