JEE Main 2023 — Mathematics Calculus

Let,

$$I={\int }_{\frac{1}{2}}^{2}\frac{{\mathrm{tan}}^{-1}x}{x}dx........(1)$$I=∫21​2​xtan−1x​dx........(1)

Now, let $x=\frac{1}{t}\Rightarrow dx=-\frac{1}{{t}^{2}}dt$x=t1​⇒dx=−t21​dt

So,

$$I=-{\int }_{2}^{\frac{1}{2}}\frac{{\mathrm{tan}}^{-1}(\frac{1}{t})}{\frac{1}{t}}\times \frac{1}{{t}^{2}}dt$$I=−∫221​​t1​tan−1(t1​)​×t21​dt

$$\Rightarrow I=-{\int }_{2}^{\frac{1}{2}}{\mathrm{tan}}^{-1}(\frac{1}{t})\times \frac{1}{t}dt$$⇒I=−∫221​​tan−1(t1​)×t1​dt

$$\Rightarrow I={\int }_{\frac{1}{2}}^{2}{\mathrm{cot}}^{-1}t\times \frac{1}{t}dt$$⇒I=∫21​2​cot−1t×t1​dt

$$\Rightarrow I={\int }_{\frac{1}{2}}^{2}{\mathrm{cot}}^{-1}x\times \frac{1}{x}dx.......(2)$$⇒I=∫21​2​cot−1x×x1​dx.......(2)

Now adding (1)&(2), we get,

$$2I={\int }_{\frac{1}{2}}^{2}\frac{1}{x}({\mathrm{cot}}^{-1}x+{\mathrm{tan}}^{-1}x)dx$$2I=∫21​2​x1​(cot−1x+tan−1x)dx

$$\Rightarrow 2I={\int }_{\frac{1}{2}}^{2}\frac{1}{x}\times \frac{\pi }{2}dx$$⇒2I=∫21​2​x1​×2π​dx

$$\Rightarrow 2I=\frac{\pi }{2}{[\mathrm{ln}x]}_{\frac{1}{2}}^{2}$$⇒2I=2π​[lnx]21​2​

$$\Rightarrow 2I=\frac{\pi }{2}[{\mathrm{log}}_{e}2-{\mathrm{log}}_{e}\frac{1}{2}]$$⇒2I=2π​[loge​2−loge​21​]

$$\Rightarrow 2I=\frac{\pi }{2}[2{\mathrm{log}}_{e}2]$$⇒2I=2π​[2loge​2]

$$\Rightarrow I=\frac{\pi }{2}{\mathrm{log}}_{e}2$$⇒I=2π​loge​2