Question Number 125873 by Mathgreat last updated on 14/Dec/20
Answered by MJS_new last updated on 15/Dec/20
![t=sin 2x =2sin x cos x =2sin x (√(1−sin^2 x)) ⇒ ⇒ sin x = { ((0≤x≤(π/4); ((√(1+t))/2)−((√(1−t))/2))),(((π/4)≤x≤(π/2); ((√(1+t))/2)+((√(1−t))/2))) :} → dx= { ((dt/(2(√(1+t))(√(1−t))))),((−(dt/(2(√(1+t))(√(1−t)))))) :} ∫_0 ^(π/4) ((sin x)/((1+(√(sin 2x)))^2 ))dx=∫_0 ^1 (dt/(4(1+(√t))^2 (√(1−t))))−∫_0 ^1 (dt/(4(1+(√t))^2 (√(1+t)))) ∫_(π/4) ^(π/2) ((sin x)/((1+(√(sin 2x)))^2 ))dx=∫_0 ^1 (dt/(4(1+(√t))^2 (√(1−t))))+∫_0 ^1 (dt/(4(1+(√t))^2 (√(1+t)))) ⇒ ∫_0 ^(π/2) ((sin x)/((1+(√(sin 2x)))^2 ))dx=∫_0 ^1 (dt/(2(1+(√t))^2 (√(1−t))))= [u=((1+(√(1−t)))/( (√t))) ⇔ t=((4u^2 )/((u^2 +1)^2 )) → dt=−((2t^(3/2) (√(1−t)))/(1+(√(1−t))))] =4∫_1 ^∞ (u/((u+1)^4 ))du=−(2/3)[((3u+1)/((u+1)^3 ))]_1 ^∞ =(1/3)](https://www.tinkutara.com/question/Q125913.png)
$${t}=\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:=\mathrm{2sin}\:{x}\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}\:=\begin{cases}{\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}};\:\frac{\sqrt{\mathrm{1}+{t}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{1}−{t}}}{\mathrm{2}}}\\{\frac{\pi}{\mathrm{4}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}};\:\frac{\sqrt{\mathrm{1}+{t}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{1}−{t}}}{\mathrm{2}}}\end{cases} \\ $$$$\rightarrow\:{dx}=\begin{cases}{\frac{{dt}}{\mathrm{2}\sqrt{\mathrm{1}+{t}}\sqrt{\mathrm{1}−{t}}}}\\{−\frac{{dt}}{\mathrm{2}\sqrt{\mathrm{1}+{t}}\sqrt{\mathrm{1}−{t}}}}\end{cases} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{\mathrm{sin}\:{x}}{\left(\mathrm{1}+\sqrt{\mathrm{sin}\:\mathrm{2}{x}}\right)^{\mathrm{2}} }{dx}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dt}}{\mathrm{4}\left(\mathrm{1}+\sqrt{{t}}\right)^{\mathrm{2}} \sqrt{\mathrm{1}−{t}}}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dt}}{\mathrm{4}\left(\mathrm{1}+\sqrt{{t}}\right)^{\mathrm{2}} \sqrt{\mathrm{1}+{t}}} \\ $$$$\underset{\frac{\pi}{\mathrm{4}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{sin}\:{x}}{\left(\mathrm{1}+\sqrt{\mathrm{sin}\:\mathrm{2}{x}}\right)^{\mathrm{2}} }{dx}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dt}}{\mathrm{4}\left(\mathrm{1}+\sqrt{{t}}\right)^{\mathrm{2}} \sqrt{\mathrm{1}−{t}}}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dt}}{\mathrm{4}\left(\mathrm{1}+\sqrt{{t}}\right)^{\mathrm{2}} \sqrt{\mathrm{1}+{t}}} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{sin}\:{x}}{\left(\mathrm{1}+\sqrt{\mathrm{sin}\:\mathrm{2}{x}}\right)^{\mathrm{2}} }{dx}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dt}}{\mathrm{2}\left(\mathrm{1}+\sqrt{{t}}\right)^{\mathrm{2}} \sqrt{\mathrm{1}−{t}}}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−{t}}}{\:\sqrt{{t}}}\:\Leftrightarrow\:{t}=\frac{\mathrm{4}{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\rightarrow\:{dt}=−\frac{\mathrm{2}{t}^{\mathrm{3}/\mathrm{2}} \sqrt{\mathrm{1}−{t}}}{\mathrm{1}+\sqrt{\mathrm{1}−{t}}}\right] \\ $$$$=\mathrm{4}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{u}}{\left({u}+\mathrm{1}\right)^{\mathrm{4}} }{du}=−\frac{\mathrm{2}}{\mathrm{3}}\left[\frac{\mathrm{3}{u}+\mathrm{1}}{\left({u}+\mathrm{1}\right)^{\mathrm{3}} }\right]_{\mathrm{1}} ^{\infty} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$