Question Number 135559 by john_santu last updated on 14/Mar/21
Commented by john_santu last updated on 14/Mar/21
$${old}\:{unswered} \\ $$
Answered by bemath last updated on 14/Mar/21
![M = ∫_0 ^(π/2) ((sin (((2x)/3)))/(tan (x))) dx let x = 3t M = ∫_0 ^(π/6) ((sin 2t)/(tan 3t)) (3dt) M = 3∫_0 ^(π/6) ((sin (3t−t))/(tan 3t)) dt M= 3∫_0 ^(π/6) ((sin 3tcos t−cos 3tsin t)/(tan 3t)) dt M=3∫_0 ^(π/6) cos 3tcos t dt−3∫_0 ^(π/6) ((cos^2 3tsin t)/(sin 3t))dt M_1 = (3/2)∫_0 ^(π/6) (cos 4t+cos 2t)dt M_1 =(3/2)[ ((sin 4t)/4)+((sin 2t)/2) ]_0 ^(π/6) M_1 = (3/2) [((√3)/8) +((√3)/4) ] = ((9(√3))/(16)). M_2 =3∫_0 ^(π/6) (((1−sin^2 3t)sin t)/(sin 3t))dt M_2 =3[ ∫_0 ^(π/6) ((sin t)/(3sin t−4sin^3 t))dt−∫_0 ^(π/2) sin 3tsin t dt ] M_2 =3[∫_0 ^(π/6) (dt/(3−4sin^2 t))+(1/2)∫_0 ^(π/6) (cos 4t−cos2 t)dt ] N = (3/2) [ ((sin 4t)/4)−((sin 2t)/2) ]_0 ^(π/6) N = (3/2) [ ((√3)/8)−((√3)/4) ]=−((3(√3))/(16)) for ∫_0 ^(π/3) (dt/(3−4sin^2 t)) apply Wierrstras subtitution let tan ((t/2))= u](https://www.tinkutara.com/question/Q135562.png)
$$\mathscr{M}\:=\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{\mathrm{sin}\:\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)}{\mathrm{tan}\:\left({x}\right)}\:{dx}\: \\ $$$${let}\:{x}\:=\:\mathrm{3}{t}\: \\ $$$$\mathscr{M}\:=\:\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \:\frac{\mathrm{sin}\:\mathrm{2}{t}}{\mathrm{tan}\:\mathrm{3}{t}}\:\left(\mathrm{3}{dt}\right) \\ $$$$\mathscr{M}\:=\:\mathrm{3}\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \:\frac{\mathrm{sin}\:\left(\mathrm{3}{t}−{t}\right)}{\mathrm{tan}\:\mathrm{3}{t}}\:{dt} \\ $$$$\mathscr{M}=\:\mathrm{3}\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \frac{\mathrm{sin}\:\mathrm{3}{t}\mathrm{cos}\:{t}−\mathrm{cos}\:\mathrm{3}{t}\mathrm{sin}\:{t}}{\mathrm{tan}\:\mathrm{3}{t}}\:{dt} \\ $$$$\mathscr{M}=\mathrm{3}\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \mathrm{cos}\:\mathrm{3}{t}\mathrm{cos}\:{t}\:{dt}−\mathrm{3}\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}{t}\mathrm{sin}\:{t}}{\mathrm{sin}\:\mathrm{3}{t}}{dt} \\ $$$$\mathscr{M}_{\mathrm{1}} =\:\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \left(\mathrm{cos}\:\mathrm{4}{t}+\mathrm{cos}\:\mathrm{2}{t}\right){dt} \\ $$$$\mathscr{M}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}\left[\:\frac{\mathrm{sin}\:\mathrm{4}{t}}{\mathrm{4}}+\frac{\mathrm{sin}\:\mathrm{2}{t}}{\mathrm{2}}\:\right]_{\mathrm{0}} ^{\pi/\mathrm{6}} \\ $$$$\mathscr{M}_{\mathrm{1}} =\:\frac{\mathrm{3}}{\mathrm{2}}\:\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\right]\:=\:\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{16}}. \\ $$$$\mathscr{M}_{\mathrm{2}} =\mathrm{3}\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \frac{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{3}{t}\right)\mathrm{sin}\:{t}}{\mathrm{sin}\:\mathrm{3}{t}}{dt} \\ $$$$\mathscr{M}_{\mathrm{2}} =\mathrm{3}\left[\:\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \frac{\mathrm{sin}\:{t}}{\mathrm{3sin}\:{t}−\mathrm{4sin}\:^{\mathrm{3}} {t}}{dt}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:\mathrm{3}{t}\mathrm{sin}\:{t}\:{dt}\:\right] \\ $$$$\mathscr{M}_{\mathrm{2}} =\mathrm{3}\left[\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \frac{{dt}}{\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} {t}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \left(\mathrm{cos}\:\mathrm{4}{t}−\mathrm{cos2}\:{t}\right){dt}\:\right] \\ $$$$\mathscr{N}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\left[\:\frac{\mathrm{sin}\:\mathrm{4}{t}}{\mathrm{4}}−\frac{\mathrm{sin}\:\mathrm{2}{t}}{\mathrm{2}}\:\right]_{\mathrm{0}} ^{\pi/\mathrm{6}} \\ $$$$\mathscr{N}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\left[\:\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\right]=−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{16}} \\ $$$${for}\:\int_{\mathrm{0}} ^{\pi/\mathrm{3}} \frac{{dt}}{\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} {t}} \\ $$$${apply}\:{Wierrstras}\:{subtitution} \\ $$$${let}\:\mathrm{tan}\:\left(\frac{{t}}{\mathrm{2}}\right)=\:{u}\: \\ $$
Answered by dhgt last updated on 07/May/21
