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

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

Answered by Frix last updated on 07/Jun/24

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

Commented by Ghisom last updated on 07/Jun/24

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

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