Question Number 179137 by yaslm last updated on 25/Oct/22
Answered by MJS_new last updated on 26/Oct/22
![∫((sin 3x)/( (√(1−sin x cos x))))dx= [t=tan x → dx=(dt/(t^2 +1))] =−∫((t(t^2 −3))/((t^2 +1)^2 (√(t^2 −t+1))))dt= [u=(((√3)(2t−1))/3) → dt=((√3)/2)du] =−2∫((3(√3)u^3 +9u^2 −9(√3)u−11)/( (√(u^2 +1))(3u^2 +2(√3)u+5)^2 ))du= [v=u+(√(u^2 +1)) → du=((√(u^2 +1))/v)dv] =−4∫((3(√3)v^6 +18v^5 −45(√3)v^4 −124v^3 +45(√3)v^2 +18v−3(√3))/((3v^4 +4(√3)v^3 +14v^2 −4(√3)v+3)^2 ))dv= [w=v+((√3)/3) → dv=dw] =−((4(√3))/3)∫((w^6 −20w^4 +((32(√3))/3)w^3 +((64)/3)w^2 −((128(√3))/9)w+((128)/(27)))/((w^2 −((2(√6))/3)w+((4(2−(√2)))/3))^2 (w^2 +((2(√6))/3)w+((4(2+(√2)))/3))^2 ))dw= [Ostrogradski′s Method] =((((4(√3))/3)w^3 +((16)/3)w^2 −((80(√3))/9)w+((32)/3))/((w^2 −((2(√6))/3)w+((4(2−(√2)))/3))(w^2 +((2(√6))/3)w+((4(2+(√2)))/3))))+((32)/3)∫((w−((√3)/3))/((w^2 −((2(√6))/3)w+((4(2−(√2)))/3))(w^2 +((2(√6))/3)w+((4(2+(√2)))/3))))dw= [decompose & formula] =((((4(√3))/3)w^3 +((16)/3)w^2 −((80(√3))/9)w+((32)/3))/((w^2 −((2(√6))/3)w+((4(2−(√2)))/3))(w^2 +((2(√6))/3)w+((4(2+(√2)))/3))))+ +((√2)/2)ln ((3w^2 −2(√6)w+4(2−(√2)))/(3w^2 +2(√6)w+4(2+(√2)))) + +(√2)arctan ((2(√2)((√3)w−2))/(3w^2 )) that′s the path, I might have made some mistakes though, it′s a bit complicated...](https://www.tinkutara.com/question/Q179158.png)
$$\int\frac{\mathrm{sin}\:\mathrm{3}{x}}{\:\sqrt{\mathrm{1}−\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=−\int\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{3}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \sqrt{{t}^{\mathrm{2}} −{t}+\mathrm{1}}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}−\mathrm{1}\right)}{\mathrm{3}}\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du}\right] \\ $$$$=−\mathrm{2}\int\frac{\mathrm{3}\sqrt{\mathrm{3}}{u}^{\mathrm{3}} +\mathrm{9}{u}^{\mathrm{2}} −\mathrm{9}\sqrt{\mathrm{3}}{u}−\mathrm{11}}{\:\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}\left(\mathrm{3}{u}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}{u}+\mathrm{5}\right)^{\mathrm{2}} }{du}= \\ $$$$\:\:\:\:\:\left[{v}={u}+\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{du}=\frac{\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}{{v}}{dv}\right] \\ $$$$=−\mathrm{4}\int\frac{\mathrm{3}\sqrt{\mathrm{3}}{v}^{\mathrm{6}} +\mathrm{18}{v}^{\mathrm{5}} −\mathrm{45}\sqrt{\mathrm{3}}{v}^{\mathrm{4}} −\mathrm{124}{v}^{\mathrm{3}} +\mathrm{45}\sqrt{\mathrm{3}}{v}^{\mathrm{2}} +\mathrm{18}{v}−\mathrm{3}\sqrt{\mathrm{3}}}{\left(\mathrm{3}{v}^{\mathrm{4}} +\mathrm{4}\sqrt{\mathrm{3}}{v}^{\mathrm{3}} +\mathrm{14}{v}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}{v}+\mathrm{3}\right)^{\mathrm{2}} }{dv}= \\ $$$$\:\:\:\:\:\left[{w}={v}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:\rightarrow\:{dv}={dw}\right] \\ $$$$=−\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\int\frac{{w}^{\mathrm{6}} −\mathrm{20}{w}^{\mathrm{4}} +\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{3}}{w}^{\mathrm{3}} +\frac{\mathrm{64}}{\mathrm{3}}{w}^{\mathrm{2}} −\frac{\mathrm{128}\sqrt{\mathrm{3}}}{\mathrm{9}}{w}+\frac{\mathrm{128}}{\mathrm{27}}}{\left({w}^{\mathrm{2}} −\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}{w}+\frac{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}{\mathrm{3}}\right)^{\mathrm{2}} \left({w}^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}{w}+\frac{\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{3}}\right)^{\mathrm{2}} }{dw}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=\frac{\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}{w}^{\mathrm{3}} +\frac{\mathrm{16}}{\mathrm{3}}{w}^{\mathrm{2}} −\frac{\mathrm{80}\sqrt{\mathrm{3}}}{\mathrm{9}}{w}+\frac{\mathrm{32}}{\mathrm{3}}}{\left({w}^{\mathrm{2}} −\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}{w}+\frac{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}{\mathrm{3}}\right)\left({w}^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}{w}+\frac{\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{3}}\right)}+\frac{\mathrm{32}}{\mathrm{3}}\int\frac{{w}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{\left({w}^{\mathrm{2}} −\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}{w}+\frac{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}{\mathrm{3}}\right)\left({w}^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}{w}+\frac{\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{3}}\right)}{dw}= \\ $$$$\:\:\:\:\:\left[\mathrm{decompose}\:\&\:\mathrm{formula}\right] \\ $$$$=\frac{\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}{w}^{\mathrm{3}} +\frac{\mathrm{16}}{\mathrm{3}}{w}^{\mathrm{2}} −\frac{\mathrm{80}\sqrt{\mathrm{3}}}{\mathrm{9}}{w}+\frac{\mathrm{32}}{\mathrm{3}}}{\left({w}^{\mathrm{2}} −\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}{w}+\frac{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}{\mathrm{3}}\right)\left({w}^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}{w}+\frac{\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{3}}\right)}+ \\ $$$$\:\:\:\:\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{3}{w}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{6}}{w}+\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}{\mathrm{3}{w}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{6}}{w}+\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}\:+ \\ $$$$\:\:\:\:\:+\sqrt{\mathrm{2}}\mathrm{arctan}\:\frac{\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{3}}{w}−\mathrm{2}\right)}{\mathrm{3}{w}^{\mathrm{2}} } \\ $$$$\mathrm{that}'\mathrm{s}\:\mathrm{the}\:\mathrm{path},\:\mathrm{I}\:\mathrm{might}\:\mathrm{have}\:\mathrm{made}\:\mathrm{some} \\ $$$$\mathrm{mistakes}\:\mathrm{though},\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{complicated}… \\ $$
Commented by MJS_new last updated on 26/Oct/22
$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is} \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\mathrm{sin}\:\mathrm{3}{x}}{\:\sqrt{\mathrm{1}−\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}}{dx}=\mathrm{2}\left(\sqrt{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:−\mathrm{1}\right) \\ $$
Commented by yaslm last updated on 26/Oct/22
great sir . thank you