Question Number 227108 by gregori last updated on 31/Dec/25
$$\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{3}}\sqrt{\mathrm{cos}\:{x}+\mathrm{1}}\:{dx}\:=? \\ $$
Answered by Frix last updated on 31/Dec/25
![=(2/3)∫_0 ^π cos x sin x (√(1+cos x)) dx =^([t=(√(1+cos x))]) =(4/3)∫_0 ^(√2) (t^4 −t^2 )dt=[(4/(15))t^5 −(4/9)t^3 ]_0 ^(√2) =((8(√2))/(45))](https://www.tinkutara.com/question/Q227109.png)
$$=\frac{\mathrm{2}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}\:\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}\:{dx}\:\overset{\left[{t}=\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}\right]} {=} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\left({t}^{\mathrm{4}} −{t}^{\mathrm{2}} \right){dt}=\left[\frac{\mathrm{4}}{\mathrm{15}}{t}^{\mathrm{5}} −\frac{\mathrm{4}}{\mathrm{9}}{t}^{\mathrm{3}} \right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} =\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{45}} \\ $$