JEE Main 2022 — Mathematics Calculus

Given,

$$I(x)=\int \frac{{\mathrm{sec}}^{2}x-2022}{{\mathrm{sin}}^{2022}x}dx$$I(x)=∫sin2022xsec2x−2022​dx

On rearranging we get,

$$\Rightarrow I(x)=\int \underset{II}{\underset{⏟}{{\mathrm{sec}}^{2}x}}\cdot \underset{I}{\underset{⏟}{{\mathrm{sin}}^{-2022}x}}dx-2022\int {\mathrm{sin}}^{-2022}xdx$$⇒I(x)=∫II⏟sec2x​​⋅I⏟sin−2022x​​dx−2022∫sin−2022xdx

Now using integration by parts we get,

$$\Rightarrow I(x)=\mathrm{tan}x.{(\mathrm{sin}x)}^{-2022}+\int (2022)\mathrm{tan}x\cdot {(\mathrm{sin}x)}^{-2023}\mathrm{cos}xdx-2022\int {(\mathrm{sin}x)}^{-2022}dx$$⇒I(x)=tanx.(sinx)−2022+∫(2022)tanx⋅(sinx)−2023cosxdx−2022∫(sinx)−2022dx

$$\Rightarrow I(x)=\mathrm{tan}x.{(\mathrm{sin}x)}^{-2022}+\int (2022)\mathrm{tan}x\cdot {(\mathrm{sin}x)}^{-2022}\frac{\mathrm{cos}x}{\mathrm{sin}x}dx-2022\int {(\mathrm{sin}x)}^{-2022}dx$$⇒I(x)=tanx.(sinx)−2022+∫(2022)tanx⋅(sinx)−2022sinxcosx​dx−2022∫(sinx)−2022dx

$$\Rightarrow I(x)=\mathrm{tan}x.{(\mathrm{sin}x)}^{-2022}+\int (2022){(\mathrm{sin}x)}^{-2022}dx-2022\int {(\mathrm{sin}x)}^{-2022}dx$$⇒I(x)=tanx.(sinx)−2022+∫(2022)(sinx)−2022dx−2022∫(sinx)−2022dx

$$\Rightarrow I(x)=(\mathrm{tan}x){(\mathrm{sin}x)}^{-2022}+C$$⇒I(x)=(tanx)(sinx)−2022+C

Now at $x=\frac{\pi }{4}$x=4π​

$$I(\frac{\pi }{4})={2}^{1011}\Rightarrow {2}^{1011}=1\times {(\frac{1}{\sqrt{2}})}^{-2022}+C\Rightarrow C=0$$I(4π​)=21011⇒21011=1×(2​1​)−2022+C⇒C=0

Hence $I(x)=\frac{\mathrm{tan}x}{{(\mathrm{sin}x)}^{2022}}$I(x)=(sinx)2022tanx​

Now finding the value of $I(\frac{\pi }{6})=\frac{1}{\sqrt{3}{(\frac{1}{2})}^{2022}}=\frac{{2}^{2022}}{\sqrt{3}}$I(6π​)=3​(21​)20221​=3​22022​

And $I(\frac{\pi }{3})=\frac{\sqrt{3}}{{(\frac{\sqrt{3}}{2})}^{2022}}=\frac{{2}^{2022}}{{(\sqrt{3})}^{2021}}=\frac{1}{{3}^{1010}}I(\frac{\pi }{6})$I(3π​)=(23​​)20223​​=(3​)202122022​=310101​I(6π​)

So, ${3}^{1010}I(\frac{\pi }{3})=I(\frac{\pi }{6})$31010I(3π​)=I(6π​)