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Knowing:

$$\phi (x)=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ \sqrt { 1-{ x }^{ 2 } \sin ^{ 2 }(t) } } } $$

I am trying to demonstrate that: $\phi (x)=\frac { 1 }{ 1+x } \phi \left( \frac { 2\sqrt { x } }{ 1+x } \right) $

Having: $\varphi (t)=arcsin\left( \frac { \left( 1+x \right) \cdot sin(t) }{ 1+x\cdot { sin }^{ 2 }(t) } \right) $

I already know that the function $\varphi$ is differentiable once, as well as its inverse bijection ${ \varphi }^{ -1 }$, and ${ \varphi }^{ ' }(t)=\frac { (1+x)\cdot \left( 1-x\cdot { sin }^{ 2 }(t) \right) }{ \left( 1+x\cdot { sin }^{ 2 }(t) \right) \sqrt { 1-{ x }^{ 2 }\cdot { sin }^{ 2 }(t) } } $

I tried to use a change of variable in $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ \sqrt { 1-{ x }^{ 2 }\cdot { sin }^{ 2 }(t) } } } $ with $t=u(t)$ but I always get stuck. If someone could help, thank you in advance.

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Do you know that your integral is an elliptic integral, found here: dlmf.nist.gov/19.2 –  Ron Gordon Jan 7 '13 at 8:55
    
Thank you, i'll try to learn more about it. –  LaX Jan 7 '13 at 8:59

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