How to prove this integral problem? For $0<k<1$ put $$K(k)=\int\limits_{0}^{\pi/2}\frac{\text{d}t}{\sqrt{1-k^2\sin^2 t}} \quad \textbf{and} \quad k'=\sqrt{1-k^2}\in(0,1)$$


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*Prove that $$K(k)=\frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$$

*Prove that $$K(k)=\frac{2}{1+k'}K\left(\frac{1-k'}{1+k'}\right)$$


I have no idea that how to solve this integral.
 A: We're given:
$$K(k) = \int_0^{\pi/2} \frac{dt}{\sqrt{1 - k^2\sin^2t}}$$
So then:
$$\frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right) = \frac{1}{1+k} \int_0^{\pi/2} \frac{dt}{\sqrt{1 - \left(\frac{2\sqrt{k}}{1+k}\right)^2\sin^2t}}$$
Simplify this and show that it equals $K(k)$.  Proceed similarly for the other problem.
A: I will show the first result using a substitution of 
$$\tan t = \dfrac{\sin 2 \theta}{k + \cos 2 \theta}.$$
Under this substitution it can be shown (it is a bit of a slog) that
$$dt = \frac{2(1 + k \cos 2\theta)}{1 + 2k \cos 2\theta + k^2} \, d\theta,$$
while the term appearing in the denominator of the integral becomes
$$1 - k^2 \sin^2 t = \frac{(1 + k \cos 2\theta)^2}{1 + 2k \cos 2\theta + k^2},$$
The limits of integration remain unchanged.
Thus
\begin{align*}
K(k) &= 2 \int^{\pi/2}_0 \frac{1 + k \cos 2\theta}{1 + 2k \cos 2 \theta + k^2} \cdot \frac{\sqrt{1 + 2k \cos 2 \theta + k^2}}{1 + k \cos 2\theta} \, d\theta\\
&= 2 \int^{\pi/2}_0 \frac{d\theta}{\sqrt{1 + 2k \cos 2\theta + k^2}}\\
&= 2 \int^{\pi/2}_0 \frac{d\theta}{\sqrt{(1 + k)^2 - 4k \sin^2 \theta}}\\
&= \frac{2}{1 + k} \int^{\pi/2}_0 \frac{d\theta}{\sqrt{1 - \left (\frac{2 \sqrt{2}}{1 + k} \right )^2 \sin^2 \theta}}, \quad {\rm since} \,\, 0 < k < 1\\
&= \frac{2}{1 + k} K \left (\frac{2 \sqrt{k}}{1 + k} \right ),
\end{align*}
as required to show. 
