Question

$\tiny \displaystyle \int \limits_{0}^{ \frac{\pi}{2} } \arccos\bigg( \frac{\cos(x) }{1 + 2 \cos(x) } \bigg)dx$​

Answer 1

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Answer 2

$\displaystyle \sf\int \limits_{0}^{ \frac{\pi}{2} } \arccos\bigg( \frac{\cos(x) }{1 + 2 \cos(x) } \bigg)dx$

$\displaystyle \sf\int \limits_{0}^{ \frac{\pi}{2} } \cos^{ - 1} \bigg( \frac{cosx }{1 + 2 cosx } \bigg)dx$

$\displaystyle \sf2\int \limits_{0}^{ { \frac{\pi}{2} } } {tan}^{ - 1} \sqrt{ \frac{ {cos}^{2} \frac{x}{2} }{2 - 3 {sin}^{2} \frac{x}{2} } } \: dx$

$\displaystyle \sf\int \limits_{0}^{ { \frac{\pi}{2} } } \int \limits_{0}^{ { 1} } \frac{\sqrt{ \frac{ {cos}^{2} \frac{x}{2} }{2 - 3 {sin}^{2} \frac{x}{2}}}}{1 + \frac{ {cos}^{2} \frac{x}{2} }{2 - 3 {sin}^{2} \frac{x}{2}} {y}^{2} } \: dydx$

$\displaystyle \sf 2 \int \limits_{0}^{ { 1 } } \int \limits_{0}^{ 1 } \frac{ \sqrt{2 - 3 {sin}^{2} \frac{x}{2} }cos \frac{x}{2} }{2 - 3 {sin}^{2} \frac{x}{2} + (1 - {sin}^{2} \frac{x}{2}) {y}^{2} } \: dxdy \left[ \sqrt{3} sin \frac{x}{2} \to \frac{ \sqrt{2} x}{ \sqrt{1 + {x}^{2} } } \right]$

$\displaystyle \sf 8 \sqrt{3} \int \limits_{0}^{ 1 }\int \limits_{0}^{ \sqrt{3} } \frac{dxdy }{ (6 + 3 {y}^{2} + {y}^{2} {x}^{2} )( {x}^{2} + 1 )} dx$

$\displaystyle \sf 4\sqrt{3} \int \limits_{0}^{ 1 } \frac{dy }{3 + {y}^{2} } \int \limits_{0}^{ \sqrt{3}}\left ( \frac{1}{ {x}^{2} + 1 } - \frac{ {y}^{2} }{ {y}^{2} {x}^{2} + 6 + 3 {y}^{2} } \right) \: dx$

$\displaystyle \sf 4 \int \limits_{0}^{ 1 } \frac{dy }{3 + {y}^{2} } \left ( \frac{\pi}{ \sqrt{3} } - \frac{ {y} }{ \sqrt{2 + {y}^{2} } } {tan}^{ - 1} \frac{y}{ \sqrt{2 + {y}^{2} } } \right)$

$\displaystyle \sf \frac{2 { \pi}^{2} }{9} - 4 \int \limits_{0}^{ 1} \left[ \frac{d}{dy} ( {tan}^{ - 1} \sqrt{2 + {y}^{2} } ){tan}^{ - 1} \frac{y}{ \sqrt{2 + {y}^{2} } } \right] \: dy$

$\displaystyle \sf \frac{2 { \pi}^{2} }{9} - 4 \left[ {tan}^{ - 1} \sqrt{2 + {y}^{2} } {tan}^{ - 1} \frac{y}{ \sqrt{2 + {y}^{2} } } \right]^{1}_{0} + 4 \int \limits^{1}_{0} \frac{ {tan}^{ - 1} \sqrt{2 + {y}^{2} } }{(1 + {y}^{2} ) \sqrt{2 + {y}^{2} } } \: dy$

$\displaystyle \sf4\int \limits_{0}^{ 1} \frac{ \frac{\pi}{2} - { tan}^{ - 1} \frac{1}{ \sqrt{2 + {y}^{2} } } }{(1 + {y}^{2} ) \sqrt{2 + {y}^{2} } } \: dy$

$\displaystyle \sf 2\pi \left[ {tan}^{ - 1} \bigg( \frac{y}{ \sqrt{2 + {y}^{2} } } \bigg)\right]^{1}_{0} - 4\int \limits_{0}^{ 1 } \int \limits_{0}^{ 1 } \frac{dxdy}{(1 + {y}^{2})(2 + {x}^{2} + {y}^{2} ) }$

$\displaystyle \frac{ {\pi}^{2} }{3} - 2\left( \sf\int \limits_{0}^{1 } \frac{dx}{1 + {x}^{2} } \right)^{2}$

$\boxed{ \therefore \displaystyle \sf\int \limits_{0}^{ \frac{\pi}{2} } \arccos\bigg( \frac{\cos(x) }{1 + 2 \cos(x) } \bigg)dx = \frac{5 {\pi}^{2} }{24} }$