Help me prove $K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$ where $K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$ Hi this problem is posed in Sean M. Stewart's How to Integrate It book pg.143
$$K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$$ where $ 0 \leq k < 1$
Based on the substitution:
$$ \tan{u} = \frac{\sin{2\theta}}{k+k\cos{2\theta}} =\frac{\tan{\theta}}{k}  $$
Show that:
$$K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$$
$K(k) $ is some function defined in term of definite-integral called complete elliptic integral which is not easily computable, so attacking $K(k)$ gave no fruitful result.
Well then proceeding with substitution ( It is easy to lose track as it  involved a lot of trigonometric manipulations )
I reduced the integral into this form :
$$K(k) = k\int^{\pi/2}_0 \frac{\sec^2{\theta}d\theta}{\sqrt{k^2 + \tan^2{\theta}}\sqrt{k^2 + \tan^2{\theta} - k^2\tan^2{\theta}}}$$
Using substitution : $\tan{\theta} = p$
$$K(k) = k\int^{\infty}_0 \frac{dp}{\sqrt{k^2 +p^2}\sqrt{k^2 +p^2 - (kp)^2}}$$
To my naive eyes this looked manageable but WolframAlpha says otherwise.
Any hint or ideas on how to attack this problem?
I thought  the trig manipulations to find a more suitable form but in the I end up transform $\sin {u}$ term into $\tan{u}$ to use the given substitution
EDIT:  It has been suggested in commens that there is typo in the substituion(given in book) , it should be :
$$ \tan{u} = \frac{\sin{2\theta}}{k+\cos{2\theta}} $$

I think that you are on the right track. But you have to be aware that the completed elliptic integral of this type can be defined by any of three integrals:
$$
K(k)=\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2\sin^2\theta}}\ d\theta\\=\int_0^1\frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}\ dt\\
=\int_0^\infty \frac{1}{\sqrt{(1+t^2)[(1+(1-k^2)t^2)}]}\ dt
$$
This is from K. Oldham, J. Myland, & J. Spanier, An Atlas of Functions, Springer . Chapter 61 covers the Complete Elliptic Integrals. Can you take it from here?
