Evaluate $$\int_0^\frac{\pi}{4}\frac{\sin x}{\cos x\sqrt{\cos 2x}} \,\rm{d}x$$
I was thinking to write $\cos 2x = \cos^2 x – \sin^2 x$ and then to factor out $\sin^2x$ to try to simplify the expression, but gets me nowhere. Any help will be appreciated!
Best Answer
\begin{align}J&=\int_0^\frac{\pi}{4}\frac{\sin x}{\cos x\sqrt{\cos 2x}} \,\rm{d}x\\ &=\int_0^\frac{\pi}{4}\frac{\sin x}{\cos x\sqrt{2\cos^2 x-1}} \,\rm{d}x\\ &\overset{t=\cos x}=\int_{\frac{1}{\sqrt{2}}}^1\frac{1}{x\sqrt{2t^2-1}}\,dt\\ &\overset{x=t^2}=\frac{1}{2}\int_{\frac{1}{2}}^1\frac{1}{x\sqrt{2x-1}}\,dx\\ &\overset{y=\sqrt{2x-1}}=\int_0^1 \frac{1}{1+y^2}\,dy\\ &=\arctan(1)-\arctan(0)\\ &=\boxed{\frac{\pi}{4}} \end{align}