The value of the integral _(5 )/(24)^( )/(24)(dx)/(1+√3 2x) is :

Question

The value of the integral ${\int }_{\frac{5\pi }{24}}^{\frac{\pi }{24}}\frac{dx}{1+\sqrt3	an 2x}$ is :

Accepted Answer

$\begin{gathered} I= {\int }_{\frac{5\pi }{24}}^{\frac{\pi }{24}}\frac{1}{1+(	an 2x{)}^{1/3}}dx\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(1) \ \Rightarrow I={\int }_{\frac{5\pi }{24}}^{\frac{\pi }{24}}\frac{1}{1+	an {(2(\frac{5\pi }{24}+\frac{\pi }{24}-4))}^{1/3}}dx \ I={\int }_{\frac{5\pi }{24}}^{\frac{\pi }{24}}\frac{1}{1+	an {(\frac{\pi }{2}-2x)}^{1/3}}dx\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(2) \ \end{gathered}$

$\begin{gathered} 2I={\int }_{\frac{5\pi }{24}}^{\frac{\pi }{24}}\frac{(	an (2x){)}^{1/3}+1}{1+(	an 2x{)}^{1/3}}dx \ ={(x)|}_{\frac{5\pi }{24}}^{\frac{\pi }{24}}\Rightarrow I=\frac{1}{2}	imes \frac{4\pi }{24}=\frac{\pi }{12} \end{gathered}$

Solutions · 1