Question

Solve: $\stackrel{\frac{\pi }{2}}{\underset{0}{\int }}\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}dx$

Answer 1

${\int }_0^{\frac{\pi }{2}}\sqrt{\left(\frac{\cos 2-\sin x}{\cos x+\sin x}\right)}dx$

$={\int }_0^{\frac{\pi }{2}}\left|\frac{\cos x-\sin x}{\cos x+\sin x}\right|dx$

$=\underset{\left(i\right)}{{\int }_0^{\frac{\pi }{2}}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)dx}+\underset{\left(ii\right)}{{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{(\sin x-\cos x)}{(\cos x+\sin x)}dx}$

$\left(i\right)$ and $\left(ii\right)$

$\cos x+\sin x=n$

$\Rightarrow \frac{dn}{dx}=\cos x-\sin x$

$\Rightarrow dn=(\cos x-\sin x)dx$

$={\int }_0^{\frac{\pi }{4}}\frac{dn}{n}-{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{dn}{n}$

$=\left[ln\right(n){]}_0^{\frac{\pi }{4}}-[ln\left(n\right){]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$

$=\left[ln\right(\cos x+\sin x){]}_0^{\frac{\pi }{4}}-[ln(\cos x+\sin x){]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$

$=ln\left(\sqrt{2}\right)4-ln\left(1\right)-ln\left(1_{+}ln\right(\sqrt{2})$

$=2ln\sqrt{2}=ln2(A-s)$