Question-125873

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

Answered by MJS_new last updated on 15/Dec/20

t = \sin 2 x = 2 \sin x \cos x = 2 \sin x \sqrt{1 - \sin^{2} x} \Rightarrow

\Rightarrow \sin x = {\begin{cases} 0 ⩽ x ⩽ \frac{π}{4}; \frac{\sqrt{1 + t}}{2} - \frac{\sqrt{1 - t}}{2} \\ \frac{π}{4} ⩽ x ⩽ \frac{π}{2}; \frac{\sqrt{1 + t}}{2} + \frac{\sqrt{1 - t}}{2} \end{cases}

\to d x = {\begin{cases} \frac{d t}{2 \sqrt{1 + t} \sqrt{1 - t}} \\ - \frac{d t}{2 \sqrt{1 + t} \sqrt{1 - t}} \end{cases}

\underset{0}{\int^{\frac{π}{4}}} \frac{\sin x}{{(1 + \sqrt{\sin 2 x})}^{2}} d x = \underset{0}{\int^{1}} \frac{d t}{4 {(1 + \sqrt{t})}^{2} \sqrt{1 - t}} - \underset{0}{\int^{1}} \frac{d t}{4 {(1 + \sqrt{t})}^{2} \sqrt{1 + t}}

\underset{\frac{π}{4}}{\int^{\frac{π}{2}}} \frac{\sin x}{{(1 + \sqrt{\sin 2 x})}^{2}} d x = \underset{0}{\int^{1}} \frac{d t}{4 {(1 + \sqrt{t})}^{2} \sqrt{1 - t}} + \underset{0}{\int^{1}} \frac{d t}{4 {(1 + \sqrt{t})}^{2} \sqrt{1 + t}}

\Rightarrow

\underset{0}{\int^{\frac{π}{2}}} \frac{\sin x}{{(1 + \sqrt{\sin 2 x})}^{2}} d x = \underset{0}{\int^{1}} \frac{d t}{2 {(1 + \sqrt{t})}^{2} \sqrt{1 - t}} =

[u = \frac{1 + \sqrt{1 - t}}{\sqrt{t}} \Leftrightarrow t = \frac{4 u^{2}}{{(u^{2} + 1)}^{2}} \to d t = - \frac{2 t^{3 / 2} \sqrt{1 - t}}{1 + \sqrt{1 - t}}]

= 4 \underset{1}{\int^{\infty}} \frac{u}{{(u + 1)}^{4}} d u = - \frac{2}{3} {[\frac{3 u + 1}{{(u + 1)}^{3}}]}_{1}^{\infty} = \frac{1}{3}

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