# The value of the definite integral

Question:

The value of the definite integral

$\int_{\pi / 24}^{5 \pi / 24} \frac{d x}{1+\sqrt[3]{\tan 2 x}} i s$

1. $\frac{\pi}{3}$

2. $\frac{\pi}{6}$

3. $\frac{\pi}{12}$

4. $\frac{\pi}{18}$

Correct Option: , 3

Solution:

Let $\mathrm{I}=\int_{\pi / 24}^{5 \pi / 24} \frac{(\cos 2 \mathrm{x})^{1 / 3}}{(\cos 2 \mathrm{x})^{1 / 3}+(\sin 2 \mathrm{x})^{1 / 3}} \mathrm{dx}$....(i)

$\Rightarrow I=\int_{\pi / 24}^{5 \pi / 24} \frac{\left(\cos \left\{2\left(\frac{\pi}{4}-x\right)\right\}\right)^{\frac{1}{3}}}{\left(\cos \left\{2\left(\frac{\pi}{4}-x\right)\right\}\right)^{\frac{1}{3}}+\left(\sin \left\{2\left(\frac{\pi}{4}-x\right)\right\}\right)^{\frac{1}{3}}} d x$

$\left\{\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right\}$

So $I=\int_{\pi / 24}^{5 \pi / 24} \frac{(\sin 2 x)^{1 / 3}}{(\sin 2 x)^{1 / 3}+(\cos 2 x)^{1 / 3}} d x$.......(ii)

Hence $2 \mathrm{I}=\int_{\pi / 24}^{5 \pi / 24} \mathrm{dx}$ $[(\mathrm{i})+(\mathrm{ii})]$

$\Rightarrow 2 I=\frac{4 \pi}{24} \Rightarrow I=\frac{\pi}{12}$