JEE Main 2021 — Mathematics Calculus

Let $\displaystyle I={\int }_{\pi /24}^{5\pi /24}\frac{dx}{1+\sqrt[3]{\mathrm{tan}2x}}$I=∫π/245π/24​1+3tan2x​dx​

$$\Rightarrow I={\int }_{\pi /24}^{5\pi /24}\frac{(\mathrm{cos}2x{)}^{1/3}}{(\mathrm{cos}2x{)}^{1/3}+(\mathrm{sin}2x{)}^{1/3}}dx...(i)$$⇒I=∫π/245π/24​(cos2x)1/3+(sin2x)1/3(cos2x)1/3​dx...(i)

$\displaystyle \Rightarrow I={\int }_{\pi /24}^{5\pi /24}\frac{{(\mathrm{cos}{2(\frac{\pi }{4}-x)})}^{\frac{1}{3}}}{{(\mathrm{cos}{2(\frac{\pi }{4}-x)})}^{\frac{1}{3}}+{(\mathrm{sin}{2(\frac{\pi }{4}-x)})}^{\frac{1}{3}}}dx$⇒I=∫π/245π/24​(cos2(4π​−x))31​+(sin2(4π​−x))31​(cos2(4π​−x))31​​dx $\displaystyle {{\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx}$∫ab​f(x)dx=∫ab​f(a+b−x)dx

So, $\displaystyle I={\int }_{\pi /24}^{5\pi /24}\frac{(\mathrm{sin}2x{)}^{1/3}}{(\mathrm{sin}2x{)}^{1/3}+(\mathrm{cos}2x{)}^{1/3}}dx...(\mathrm{ii})$I=∫π/245π/24​(sin2x)1/3+(cos2x)1/3(sin2x)1/3​dx...(ii)

Adding the equation (i)+(ii), we get

Hence, $\displaystyle 2I={\int }_{\pi /24}^{5\pi /24}dx$2I=∫π/245π/24​dx

$$\Rightarrow 2I={[x]}_{\frac{\pi }{24}}^{\frac{5\pi }{24}}$$⇒2I=[x]24π​245π​​

$$\Rightarrow 2I=[\frac{5\pi }{24}-\frac{\pi }{24}]$$⇒2I=[245π​−24π​]

$$\Rightarrow 2I=\frac{4\pi }{24}\Rightarrow I=\frac{\pi }{12}$$⇒2I=244π​⇒I=12π​