Question

# $$\displaystyle \int_{0}^{1}\frac{dx}{(1+x^{2})\sqrt{(2+x^{2})}}=\frac{\pi }{k}$$. Find the value of $$k$$.

Solution

## Let $$\displaystyle I=\int { \frac { 1 }{ \left( { x }^{ 2 }+1 \right) \sqrt { { x }^{ 2 }+2 } } dx }$$Put $$x=\sqrt { 2 } \tan { t } \Rightarrow dx=\sqrt { 2 } \sec ^{ 2 }{ t } dt$$$$\displaystyle I=\sqrt { 2 } \int { \frac { sec{ t } }{ \sqrt { 2 } \left( 2\tan ^{ 2 }{ t } +1 \right) } dt } =\int { \frac { sec{ t } }{ 2\tan ^{ 2 }{ t } +1 } dt }$$Multiplying numerator and denominator by $$\cos ^{ 2 }{ t }$$, we get$$\displaystyle I=\int { \frac { \cos { t } }{ 2\sin ^{ 2 }{ t } +\cos ^{ 2 }{ t } } dt } =\int { \frac { \cos { t } }{ \sin ^{ 2 }{ t } +1 } dt }$$Now put $$u=\sin { t } \Rightarrow du=\cos { t } dt$$Therefore$$\displaystyle I=\int { \frac { 1 }{ { u }^{ 2 }+1 } du } =\tan ^{ -1 }{ u } =\tan ^{ -1 }{ \sin { t } }$$$$\displaystyle =\tan ^{ -1 }{ \sin { \left( \tan ^{ -1 }{ \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } } =\tan ^{ -1 }{ \left( \frac { x }{ \sqrt { { x }^{ 2 }+2 } } \right) }$$Hence$$\displaystyle \int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( { x }^{ 2 }+1 \right) \sqrt { { x }^{ 2 }+2 } } dx } ={ \left[ \tan ^{ -1 }{ \left( \frac { x }{ \sqrt { { x }^{ 2 }+2 } } \right) } \right] }_{ 0 }^{ 1 }$$$$\displaystyle =\frac { \pi }{ 6 } -0=\frac { \pi }{ 6 }$$$$\Rightarrow k=6$$Mathematics

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