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Question Number 209540 by mokys last updated on 13/Jul/24
Answered by Spillover last updated on 13/Jul/24
∫(1/(sin^6 x+sin^5 x+sin^4 x+sin^3 x+sin^2 x+sin x)) sin^6 x+sin^5 x+sin^4 x+sin^3 x+sin^2 x+sin x=Σ_(k=1) ^6 sin^k x t=tan (x/2) sin x=((2t)/(1+t^2 )) dx=((2dt)/(1+t^2 )) Σ_(k=1) ^6 sin^k x=Σ_(k=1) ^6 (((2t)/(1+t^2 )))^k ∫(1/(sin^6 x+sin^5 x+sin^4 x+sin^3 x+sin^2 x+sin x))dx ∫(1/(Σ_(k=1) ^6 (((2t)/(1+t^2 )))^k )).((2dt)/(1+t^2 ))=∫(2/(Σ_(k=1) ^6 ((2^k t^k )/((1+t^2 )^k )).(1+t^2 )))dt =∫(2/(Σ_(k=1) ^6 ((2^k t^k )/((1+t^2 )^k )).(1+t^2 )))dt=∫(2/(Σ_(k=1) ^6 ((2^k t^k )/((1+t^2 )^(k−1) ))))dt ∫(2/(Σ_(k=1) ^6 ((2^k t^k )/((1+t^2 )^(k−1) ))))dt ....
$$\int\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{sin}\:^{\mathrm{5}} {x}+\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:{x}} \\ $$$$\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{sin}\:^{\mathrm{5}} {x}+\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:{x}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\mathrm{sin}\:^{{k}} {x} \\ $$$${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\:\:\:\:\:\mathrm{sin}\:{x}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\:{dx}=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\mathrm{sin}\:^{{k}} {x}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{{k}} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{sin}\:^{\mathrm{5}} {x}+\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:{x}}{dx} \\ $$$$\int\frac{\mathrm{1}}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{{k}} }.\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\int\frac{\mathrm{2}}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\:\frac{\mathrm{2}^{{k}} {t}^{{k}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{k}} }.\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\int\frac{\mathrm{2}}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\:\frac{\mathrm{2}^{{k}} {t}^{{k}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{k}} }.\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}=\int\frac{\mathrm{2}}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\:\frac{\mathrm{2}^{{k}} {t}^{{k}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{k}−\mathrm{1}} }}{dt} \\ $$$$\int\frac{\mathrm{2}}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\:\frac{\mathrm{2}^{{k}} {t}^{{k}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{k}−\mathrm{1}} }}{dt} \\ $$$$…. \\ $$
Commented by mokys last updated on 14/Jul/24
its very hard
$${its}\:{very}\:{hard} \\ $$
Commented by Spillover last updated on 14/Jul/24
of course.
$${of}\:{course}. \\ $$
Answered by Ghisom last updated on 14/Jul/24
∫(dx/(Σ_(n=1) ^6 sin^n x))= =∫(dx/(sin x (1+sin x)(1+sin x +sin^2 x)(1−sin x +sin^2 x)))= =∫(dx/(sin x))− [=I_1 ] −(1/3)∫(dx/(1+sin x))− [=I_2 ] −(1/2)∫((1+sin x)/(1+sin x +sin^2 x))dx− [=I_3 ] −(1/6)∫((1+sin x)/(1−sin x +sin^2 x))dx [=I_4 ] I_1 =ln ∣((1−cos x)/(sin x))∣ I_2 =((1−sin x)/(3cos x)) I_3 , I_4 are hard but possible
$$\int\frac{{dx}}{\underset{{n}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\mathrm{sin}^{{n}} \:{x}}= \\ $$$$=\int\frac{{dx}}{\mathrm{sin}\:{x}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\left(\mathrm{1}+\mathrm{sin}\:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}\right)}= \\ $$$$=\int\frac{{dx}}{\mathrm{sin}\:{x}}−\:\:\:\:\:\left[={I}_{\mathrm{1}} \right] \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}}−\:\:\:\:\:\left[={I}_{\mathrm{2}} \right] \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}−\:\:\:\:\:\left[={I}_{\mathrm{3}} \right] \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}\:\:\:\:\:\left[={I}_{\mathrm{4}} \right] \\ $$$${I}_{\mathrm{1}} =\mathrm{ln}\:\mid\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\mid \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{3cos}\:{x}} \\ $$$${I}_{\mathrm{3}} ,\:{I}_{\mathrm{4}} \:\mathrm{are}\:\mathrm{hard}\:\mathrm{but}\:\mathrm{possible} \\ $$
Answered by Frix last updated on 16/Jul/24
∫(dx/(Σ_(k=1) ^6 sin^k x)) =^(t=x−(π/2)) ∫(dt/(Σ_(k=1) ^6 cos^k x)) =^(u=tan (t/2)) =−∫(((u^2 +1)^5 )/((u−1)(u+1)(u^4 +3)(3u^4 +1)))du= =−(1/3)∫du− −∫(du/(u−1))+ +∫(du/(u+1))− −2∫(du/(u^4 +3))− −(2/9)∫(du/(u^4 +(1/3))) Where for b>0: a∫(du/(u^4 +b))= =(a/(2^(5/2) b^(3/4) ))(ln ((u^2 +2^(1/2) b^(1/4) u+b^(1/2) )/(u^2 −2^(1/2) b^(1/4) u+b^(1/2) )) −2tan^(−1) ((2^(1/2) b^(1/4) u)/(u^2 −b^(1/2) )))
$$\int\frac{{dx}}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\:\mathrm{sin}^{{k}} \:{x}}\:\overset{{t}={x}−\frac{\pi}{\mathrm{2}}} {=}\:\int\frac{{dt}}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\mathrm{cos}^{{k}} \:{x}}\:\overset{{u}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}} {=} \\ $$$$=−\int\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{5}} }{\left({u}−\mathrm{1}\right)\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{4}} +\mathrm{3}\right)\left(\mathrm{3}{u}^{\mathrm{4}} +\mathrm{1}\right)}{du}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\int{du}− \\ $$$$−\int\frac{{du}}{{u}−\mathrm{1}}+ \\ $$$$+\int\frac{{du}}{{u}+\mathrm{1}}− \\ $$$$−\mathrm{2}\int\frac{{du}}{{u}^{\mathrm{4}} +\mathrm{3}}− \\ $$$$−\frac{\mathrm{2}}{\mathrm{9}}\int\frac{{du}}{{u}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\mathrm{Where}\:\mathrm{for}\:{b}>\mathrm{0}: \\ $$$${a}\int\frac{{du}}{{u}^{\mathrm{4}} +{b}}= \\ $$$$=\frac{{a}}{\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{2}}} {b}^{\frac{\mathrm{3}}{\mathrm{4}}} }\left(\mathrm{ln}\:\frac{{u}^{\mathrm{2}} +\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} {b}^{\frac{\mathrm{1}}{\mathrm{4}}} {u}+{b}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{u}^{\mathrm{2}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} {b}^{\frac{\mathrm{1}}{\mathrm{4}}} {u}+{b}^{\frac{\mathrm{1}}{\mathrm{2}}} }\:−\mathrm{2tan}^{−\mathrm{1}} \:\frac{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} {b}^{\frac{\mathrm{1}}{\mathrm{4}}} {u}}{{u}^{\mathrm{2}} −{b}^{\frac{\mathrm{1}}{\mathrm{2}}} }\right) \\ $$
Commented by Spillover last updated on 30/Jul/24
perfect
$${perfect} \\ $$

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