JEE Main 2023 — Mathematics Calculus

Let,

$$I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{x+\frac{\pi }{4}}{2-\mathrm{cos}2x}dx\ldots (1)$$I=∫4−π​4π​​2−cos2xx+4π​​dx…(1)

Now replacing $x\rightarrow -x$x→−x we get,

$$\Rightarrow I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{-x+\frac{\pi }{4}}{2-\mathrm{cos}2x}dx\ldots (2)$$⇒I=∫4−π​4π​​2−cos2x−x+4π​​dx…(2)

Now both the equation we get,

$$\Rightarrow 2I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{\frac{\pi }{2}}{2-\mathrm{cos}2x}dx$$⇒2I=∫4−π​4π​​2−cos2x2π​​dx

$$\Rightarrow I=\frac{\pi }{4}\cdot 2{\int }_{0}^{\frac{\pi }{4}}\frac{dx}{2-\mathrm{cos}2x}dx{\text{as}\mathrm{cos}2x\text{is even function}}$$⇒I=4π​⋅2∫04π​​2−cos2xdx​dxascos2xis even function

$$\Rightarrow I=\frac{\pi }{4}\cdot 2{\int }_{0}^{\frac{\pi }{4}}\frac{(1+{\mathrm{tan}}^{2}x)\mathrm{dx}}{2(1+{\mathrm{tan}}^{2}x)-(1-{\mathrm{tan}}^{2}x)}$$⇒I=4π​⋅2∫04π​​2(1+tan2x)−(1−tan2x)(1+tan2x)dx​

$$\Rightarrow I=\frac{\pi }{4}\cdot 2{\int }_{0}^{\frac{\pi }{4}}\frac{{\mathrm{sec}}^{2}x\mathrm{dx}}{3{\mathrm{tan}}^{2}x+1}$$⇒I=4π​⋅2∫04π​​3tan2x+1sec2xdx​

Now let $\mathrm{tan}x=t\Rightarrow {\mathrm{sec}}^{2}xdx=dt$tanx=t⇒sec2xdx=dt

$$\Rightarrow I=\frac{\pi }{2}{\int }_{0}^{1}\frac{dt}{3{t}^{2}+1}$$⇒I=2π​∫01​3t2+1dt​

$$\Rightarrow I=\frac{\pi }{2\sqrt{3}}{\mathrm{tan}}^{-1}\sqrt{3}$$⇒I=23​π​tan−13​

$$\Rightarrow I=\frac{{\pi }^{2}}{6\sqrt{3}}$$⇒I=63​π2​