Question

The value of the definite integral $\underset{\pi /24}{\overset{5\pi /24}{\int }}\frac{dx}{1+\sqrt[3]{3\tan 2x}}$ is

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

Answer 1

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

Let $I=\underset{\pi /24}{\overset{5\pi /24}{\int }}\frac{dx}{1+(\tan 2x{)}^{1/3}}\dots \left(1\right)$
applying ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$,
we get
$I=\underset{\pi /24}{\overset{5\pi /24}{\int }}\frac{dx}{1+{\left(\tan (\frac{\pi }{2}-2x)\right)}^{1/3}}\Rightarrow I=\underset{\pi /24}{\overset{5\pi /24}{\int }}\frac{dx}{1+(\cot 2x{)}^{1/3}}\Rightarrow I=\underset{\pi /24}{\overset{5\pi /24}{\int }}\frac{(\tan 2x{)}^{1/3}}{1+(\tan 2x{)}^{1/3}}dx\dots \left(2\right)$

By $\left(1\right)+\left(2\right)$
$2I=\underset{\pi /24}{\overset{5\pi /24}{\int }}\frac{1+(\tan 2x{)}^{1/3}}{1+(\tan 2x{)}^{1/3}}dx\Rightarrow \underset{\pi /24}{\overset{5\pi /24}{\int }}1dx\Rightarrow 2I=\frac{5\pi }{24}-\frac{\pi }{24}\Rightarrow I=\frac{\pi }{12}$