Evaluate the definite integral: $\int\limits_0^{\pi/2}\frac{\sin x-\cos x}{\sqrt{1-\sin 2x}}\, dx$

Question

I would like to evaluate :

$$\int\limits_0^{\pi/2}\frac{\sin x-\cos x}{\sqrt{1-\sin 2x}}\, dx$$

progress

$I=\int\limits_0^{\pi/2}\frac{\sin x-\cos x}{\sqrt{1-\sin 2x}}\, dx$
$=\int\limits_0^{\pi/2}\frac{\sin x-\cos x}{\sqrt{(\sin x-\cos x)^2}}\, dx$
$=\pi/2$

I am not sure whether my answer is correct or not.
I want to sure about this, please someone confirmed me.

Answer 1

Note that $$\int\limits_0^{\pi/2}\frac{\sin x-\cos x}{\sqrt{(\sin x-\cos x)^2}}\, dx= \int\limits_0^{\pi/2}\frac{\sin x-\cos x}{|\sin x-\cos x|}\, dx = \int\limits_{-\pi/4}^{\pi/4}\frac{\sin(u)}{|\sin(u)|}\, du=0.$$

For the second equality we use the substitution $u=x-\pi/4$ and the fact that $$\sin(u+\pi/4)-\cos(u+\pi/4)=\sqrt{2}\sin(u).$$ For the last equality, remark that you're integrating an odd function on a symetric interval.

Answer 2

$$\sin x +\cos x =t$$ $$\implies (\cos x -\sin x)dx=dt$$ $$2\sin x\cos x=t^2 -1 $$ $$\int \frac{-dt}{\sqrt{2-t^2}}$$

Answer 3

It should be easy now to find the solution to come out to be 0

Answer 4

Take care of modulus function.....