How to prove the following identity? For $0\le m<1$, $$\int_{0}^{\pi/2}\frac{dx}{\sqrt{(1-m \cos^2 x)(1+ m \sin^2x)}}=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-m^2 \sin^2 x}}$$ Then it will be an elliptic integral.
Firstly, the objects in the squre root are not equal, so it cann't be solved just by Trigonometric Identities.
Let $$I(a, b) =\int_{0}^{\pi/2}\frac{dx}{\sqrt{a^{2}\cos^{2}x+b^{2}\sin^{2}x}}\tag{1}$$ then it is easy to observe that $$cI(ac, bc)=I(a, b) \tag{2}$$ Using the substitution $b\tan x=t$ in $(1)$ we see that $$I(a, b) =\int_{0}^{\infty}\frac{dt}{\sqrt{(t^{2}+a^{2})(t^{2}+b^{2})}}\tag{3}$$ The integral on the right hand side of the identity in question is a complete elliptic integral of first kind and is commonly denoted by $K(m)$. It is almost obvious from equation $(1)$ that $$K(m) =I(1, \sqrt{1-m^{2}})\tag{4}$$ Using substitution $\tan x=t$ we can see that the left hand side of the identity in question is equal to $$\int_{0}^{\infty}\frac{dt}{\sqrt{(t^{2}+1-m)(1+(1+m)t^{2})}}=\frac{I(1/\sqrt{1+m},\sqrt{1-m})}{\sqrt{1+m}}=I(1,\sqrt{1-m^{2}})$$ (via equations $(2)$ and $(3)$). The proof is now complete via equation $(4)$.