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Question Number 125873 by Mathgreat last updated on 14/Dec/20

Answered by MJS_new last updated on 15/Dec/20

$$t=\sin 2x=2\sin x\cos x=2\sin x\sqrt{1-{\sin }^{2}x}\Rightarrow$$$$\Rightarrow \sin x=\{\begin{array}{l}0⩽x⩽\frac{\pi }{4};\frac{\sqrt{1+t}}{2}-\frac{\sqrt{1-t}}{2}\\ \frac{\pi }{4}⩽x⩽\frac{\pi }{2};\frac{\sqrt{1+t}}{2}+\frac{\sqrt{1-t}}{2}\end{array}$$$$\to dx=\{\begin{array}{l}\frac{dt}{2\sqrt{1+t}\sqrt{1-t}}\\ -\frac{dt}{2\sqrt{1+t}\sqrt{1-t}}\end{array}$$$$\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}\frac{\sin x}{{(1+\sqrt{\sin 2x})}^2}dx=\underset{0}{\stackrel{1}{\int }}\frac{dt}{4{(1+\sqrt{t})}^2\sqrt{1-t}}-\underset{0}{\stackrel{1}{\int }}\frac{dt}{4{(1+\sqrt{t})}^2\sqrt{1+t}}$$$$\underset{\frac{\pi }{4}}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\sin x}{{(1+\sqrt{\sin 2x})}^2}dx=\underset{0}{\stackrel{1}{\int }}\frac{dt}{4{(1+\sqrt{t})}^2\sqrt{1-t}}+\underset{0}{\stackrel{1}{\int }}\frac{dt}{4{(1+\sqrt{t})}^2\sqrt{1+t}}$$$$\Rightarrow$$$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\sin x}{{(1+\sqrt{\sin 2x})}^2}dx=\underset{0}{\stackrel{1}{\int }}\frac{dt}{2{(1+\sqrt{t})}^2\sqrt{1-t}}=$$$$[u=\frac{1+\sqrt{1-t}}{\sqrt{t}}\iff t=\frac{4u^2}{{(u^2+1)}^2}\to dt=-\frac{2t^{3/2}\sqrt{1-t}}{1+\sqrt{1-t}}]$$$$=4\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{u}{{(u+1)}^4}du=-\frac{2}{3}{\left[\frac{3u+1}{{(u+1)}^3}\right]}_1^{\mathrm{\infty }}=\frac{1}{3}$$

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