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Alignment and spacing. In particular, the incorrectly small space between 2 and \tan resulted from seemingly purposeless {curly braces} enclosing \tan\frac z 2.
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edit approved Jun 17, 2017 at 17:52
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edit approved Jun 17, 2017 at 17:52
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We have $$2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}}dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\\ =\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}}dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}}=\\ 2\int_0^{\pi/2}\frac{\cos z}{1+\cos z}dz=\pi-2\int_0^{\pi/2}\frac1{1+\cos z}dz=\pi-2{\tan\frac{z}2\bigg|}_{0}^{\pi/2}=\pi-2,$$

\begin{align} & 2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}} \, dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}} \, dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy \\[6pt] = {} &\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy =\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}} \, dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}} \\[6pt] = {} &2\int_0^{\pi/2}\frac{\cos z}{1+\cos z} \, dz=\pi-2\int_0^{\pi/2}\frac1{1+\cos z} \,dz= \pi-2\tan\frac{z}2\bigg|_0^{\pi/2}=\pi-2, \end{align} where we used substitutions $y=2x$, $t=\sin y$, $t=\cos^2 z$.

We have $$2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}}dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\\ =\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}}dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}}=\\ 2\int_0^{\pi/2}\frac{\cos z}{1+\cos z}dz=\pi-2\int_0^{\pi/2}\frac1{1+\cos z}dz=\pi-2{\tan\frac{z}2\bigg|}_{0}^{\pi/2}=\pi-2,$$ where we used substitutions $y=2x$, $t=\sin y$, $t=\cos^2 z$.

We have

\begin{align} & 2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}} \, dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}} \, dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy \\[6pt] = {} &\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy =\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}} \, dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}} \\[6pt] = {} &2\int_0^{\pi/2}\frac{\cos z}{1+\cos z} \, dz=\pi-2\int_0^{\pi/2}\frac1{1+\cos z} \,dz= \pi-2\tan\frac{z}2\bigg|_0^{\pi/2}=\pi-2, \end{align} where we used substitutions $y=2x$, $t=\sin y$, $t=\cos^2 z$.

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Fedor Petrov
Fedor Petrov
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We have $$2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}}dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\\ =\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}}dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}}=\\ 2\int_0^{\pi/2}\frac{\cos z}{1+\cos z}dz=\pi-2,$$$$2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}}dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\\ =\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}}dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}}=\\ 2\int_0^{\pi/2}\frac{\cos z}{1+\cos z}dz=\pi-2\int_0^{\pi/2}\frac1{1+\cos z}dz=\pi-2{\tan\frac{z}2\bigg|}_{0}^{\pi/2}=\pi-2,$$ where we used substitutions $y=2x$, $t=\sin y$, $t=\cos^2 z$.

We have $$2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}}dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\\ =\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}}dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}}=\\ 2\int_0^{\pi/2}\frac{\cos z}{1+\cos z}dz=\pi-2,$$ where we used substitutions $y=2x$, $t=\sin y$, $t=\cos^2 z$.

We have $$2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}}dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\\ =\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}}dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}}=\\ 2\int_0^{\pi/2}\frac{\cos z}{1+\cos z}dz=\pi-2\int_0^{\pi/2}\frac1{1+\cos z}dz=\pi-2{\tan\frac{z}2\bigg|}_{0}^{\pi/2}=\pi-2,$$ where we used substitutions $y=2x$, $t=\sin y$, $t=\cos^2 z$.

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Fedor Petrov
Fedor Petrov
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We have $$2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}}dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\\ =\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}}dy=\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}}dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}}=\\ 2\int_0^{\pi/2}\frac{\cos z}{1+\cos z}dz=\pi-2,$$ where we used substitutions $y=2x$, $t=\sin y$, $t=\cos^2 z$.