JEE Main 2019 — Mathematics Calculus

Let $\displaystyle I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\mathrm{sec}}^{\frac{2}{3}}x\cdot {\mathrm{cosec}}^{\frac{4}{3}}xdx$I=∫6π​3π​​sec32​x⋅cosec34​xdx

$$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}(\frac{1}{{\mathrm{cos}}^{\frac{2}{3}}x\cdot {\mathrm{sin}}^{\frac{4}{3}}x})dx$$⇒I=∫6π​3π​​(cos32​x⋅sin34​x1​)dx

$$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}(\frac{{\mathrm{cos}}^{\frac{4}{3}}x}{{\mathrm{cos}}^{\frac{2}{3}+\frac{4}{3}}x\cdot {\mathrm{sin}}^{\frac{4}{3}}x})dx$$⇒I=∫6π​3π​​(cos32​+34​x⋅sin34​xcos34​x​)dx

$$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}(\frac{{\mathrm{sec}}^{2}x}{{\mathrm{tan}}^{\frac{4}{3}}x})dx$$⇒I=∫6π​3π​​(tan34​xsec2x​)dx

Let $\mathrm{tan}x=t,{ sec}^{2}xdx=dt$tanx=t,sec2xdx=dt and at $x=\frac{\pi }{6},t=\mathrm{tan}(\frac{\pi }{6})=\frac{1}{\sqrt{3}}$x=6π​,t=tan(6π​)=3​1​ and at $x=\frac{\pi }{3},t=\mathrm{tan}(\frac{\pi }{3})=\sqrt{3}$x=3π​,t=tan(3π​)=3​

$$\Rightarrow I={\int }_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{dt}{{t}^{\frac{4}{3}}}$$⇒I=∫3​1​3​​t34​dt​

Using $\displaystyle \int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}$∫xndx=n+1xn+1​

$$\Rightarrow I={[\frac{{t}^{-\frac{4}{3}+1}}{-\frac{4}{3}+1}]}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$$⇒I=[−34​+1t−34​+1​]3​1​3​​

$$\Rightarrow I={[-3{t}^{-\frac{1}{3}}]}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$$⇒I=[−3t−31​]3​1​3​​

$$\Rightarrow I=-3{{(\sqrt{3})}^{-\frac{1}{3}}-{(\frac{1}{\sqrt{3}})}^{-\frac{1}{3}}}$$⇒I=−3(3​)−31​−(3​1​)−31​

$$\Rightarrow I=-3{{({3}^{\frac{1}{2}})}^{-\frac{1}{3}}-{(\frac{1}{{3}^{\frac{1}{2}}})}^{-\frac{1}{3}}}$$⇒I=−3(321​)−31​−(321​1​)−31​

$$=-3(\frac{1}{{3}^{\frac{1}{6}}}-{3}^{\frac{1}{6}})$$=−3(361​1​−361​)

$$=\frac{-3}{{3}^{\frac{1}{6}}}+3\times {3}^{\frac{1}{6}}$$=361​−3​+3×361​

$${=3}^{\frac{7}{6}}-{3}^{\frac{5}{6}}.$$=367​−365​.