Question

The integral ${\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}\frac{dx}{1+\sqrt[3]{3\tan 2x}}$ is equal to:

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{18}$

Answer 1

The correct option is B $\frac{\pi }{12}$

$I={\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}\frac{dx}{1+\sqrt[3]{3\tan 2x}}$ ... (1)

Also,

$I={\int }_{\frac{\pi }{4}-\frac{\pi }{24}}^{\frac{\pi }{4}-\frac{5\pi }{24}}\frac{dx}{1+\sqrt[3]{3\tan 2(\frac{\pi }{4}-x)}}$

$\therefore I={\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}\frac{dx}{1+\sqrt[3]{3\cot 2x}}$ ... (2)

Adding (1) and (2), we have

$2I={\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}\frac{dx}{1+\sqrt[3]{3\cot 2x}}+\frac{dx}{1+\sqrt[3]{3\tan 2x}}$

$\therefore 2I={\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}\frac{2+\sqrt[3]{3\cot 2x}+\sqrt[3]{3\tan 2x}}{(1+\sqrt[3]{3\cot 2x})(1+\sqrt[3]{3\tan 2x})}dx$

$\therefore 2I={\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}\frac{2+\sqrt[3]{3\cot 2x}+\sqrt[3]{3\tan 2x}}{2+\sqrt[3]{3\cot 2x}+\sqrt[3]{3\tan 2x}}dx$

$\therefore 2I={\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}dx$

$\therefore 2I=\frac{\pi }{6}$

$\therefore I=\frac{\pi }{12}$

This is required answer.