Find the mistake in evaluation of $\int_0^{\pi/2} \frac{1}{\sqrt{\sin x (2+\cos(2x)}} dx$

Question

Evaluate $$\int_0^{\pi/2} \frac{1}{\sqrt{\sin x (2+\cos(2x)}} dx$$ It is $$\frac{1}{\sqrt{\sin x (2+\cos(2x)}}=\frac{1}{\sqrt{\sin x (3-2 \sin^2 x)}}$$ So letting $\sin x=y$ it is $dx=\frac{1}{\sqrt{1-y^2}}dy$, so $$\int_0^{\pi/2} \frac{1}{\sqrt{\sin x (3-2 \sin^2 x)}} dx=\int_0^1 \frac{1}{\sqrt{1-y^2}\sqrt{y}\sqrt{3-2y^2}}dy$$ Letting $y^2=z$ it is $2ydy=dz$ and so $$\int_0^1 \frac{1}{\sqrt{1-y^2}\sqrt{y}\sqrt{3-2y^2}}dy=2\int_0^1\frac{1}{\sqrt{1-z}\sqrt{3-2z^2}}dz=2\int_0^1\frac{1}{\sqrt{3-5z+2z^2}}dz=$$ $$=4\sqrt{2}\int_0^1 \frac{1}{\sqrt{16z^2-40z+24}} dz=4\sqrt{2}\int_0^1 \frac{1}{\sqrt{(4z-5)^2-1}} dz$$ Let $4z-5=s$, then $4dz=ds$; so $$4\sqrt{2}\int_0^1 \frac{1}{\sqrt{(4z-5)^2-1}} dz=\sqrt{2}\int_{-5}^{-1} \frac{1}{\sqrt{s^2-1}}ds$$ Letting $s=\cosh u$ it is $ds=\sinh u du$ and so $$\sqrt{2}\int_{-5}^{-1} \frac{1}{\sqrt{s^2-1}}ds=\sqrt{2}\int_{\cosh^{-1}(-5)}^{\cosh^{-1}(-1)} du=\sqrt{2}[\cosh^{-1} (-1)-\cosh^{-1} (-5)]$$ The problem is that the domain of $\cosh^{-1}x$ is $x\geq1$. Where is my mistake? I notice that if I write $(4z-5)^2=(5-4z)^2$ this problem doesn't occour, why? Thanks.

Answer 1

Your mistake is the following:

The correct result of the substitution $y^2=z$ ($dy=\frac{dz}{2\sqrt z}$) is: $$\int_0^1 \frac{1}{\sqrt{1-y^2}\sqrt{y}\sqrt{3-2y^2}}dy=\int_0^1\frac{1}{\sqrt{1-z}\sqrt{3-2z}}\frac{dz}{\color{red}{2z^{3/4}}}.$$

You should also be aware that this integral cannot be evaluated in terms of elementary functions.