Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 208140 by Ghisom last updated on 06/Jun/24

∫_0 ^π (dx/(1+((sin x))^(1/3) ))=? exact result required

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dx}}{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{sin}\:{x}}}=? \\ $$$$\mathrm{exact}\:\mathrm{result}\:\mathrm{required} \\ $$

Answered by Frix last updated on 07/Jun/24

=2∫_0 ^(π/2) (dx/(1+sin^(1/3) x))=2∫_0 ^(π/2) ((1−sin^(1/3) x +sin^(2/3) x)/(1+sin x))dx= =2∫_0 ^(π/2) (((1−sin^(1/3) x +sin^(2/3) x)(1−sin x))/(cos^2 x))dx= =2∫_0 ^(π/2) ((1−sin x)/(cos^2 x))dx− −2∫_0 ^(π/2) sin^(1/3) x cos^(−2) x dx+ +2∫_0 ^(π/2) sin^(2/3) x cos^(−2) x dx+ +2∫_0 ^(π/2) sin^(4/3) x cos^(−2) x dx− −2∫_0 ^(π/2) sin^(5/3) x cos^(−2) x dx= The first one is trivial and for the others use the Beta function: B (p, q) =2∫_0 ^(π/2) sin^(2p−1) x cos^(2q−1) x dx=((Γ (p) Γ (q))/(Γ (p+q)))

$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{3}}} \:{x}}=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}−\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{3}}} \:{x}\:+\mathrm{sin}^{\frac{\mathrm{2}}{\mathrm{3}}} \:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dx}= \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\left(\mathrm{1}−\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{3}}} \:{x}\:+\mathrm{sin}^{\frac{\mathrm{2}}{\mathrm{3}}} \:{x}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}− \\ $$$$−\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{3}}} \:{x}\:\mathrm{cos}^{−\mathrm{2}} \:{x}\:{dx}+ \\ $$$$+\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}^{\frac{\mathrm{2}}{\mathrm{3}}} \:{x}\:\mathrm{cos}^{−\mathrm{2}} \:{x}\:{dx}+ \\ $$$$+\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}^{\frac{\mathrm{4}}{\mathrm{3}}} \:{x}\:\mathrm{cos}^{−\mathrm{2}} \:{x}\:{dx}− \\ $$$$−\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}^{\frac{\mathrm{5}}{\mathrm{3}}} \:{x}\:\mathrm{cos}^{−\mathrm{2}} \:{x}\:{dx}= \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{one}\:\mathrm{is}\:\mathrm{trivial}\:\mathrm{and}\:\mathrm{for}\:\mathrm{the}\:\mathrm{others} \\ $$$$\mathrm{use}\:\mathrm{the}\:\mathrm{Beta}\:\mathrm{function}: \\ $$$$\mathrm{B}\:\left({p},\:{q}\right)\:=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}^{\mathrm{2}{p}−\mathrm{1}} \:{x}\:\mathrm{cos}^{\mathrm{2}{q}−\mathrm{1}} \:{x}\:{dx}=\frac{\Gamma\:\left({p}\right)\:\Gamma\:\left({q}\right)}{\Gamma\:\left({p}+{q}\right)} \\ $$

Commented by Ghisom last updated on 07/Jun/24

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com