JEE Main 2024 — Mathematics Calculus

Given,

$${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(\frac{{x}^{2}\mathrm{cos}x}{1+{\pi }^{x}}+\frac{1+{\mathrm{sin}}^{2}x}{1+{e}^{{\mathrm{sin}}^{2023}x}})dx=\frac{\pi }{4}(\pi +\alpha )-4$$∫2−π​2π​​(1+πxx2cosx​+1+esin2023x1+sin2x​)dx=4π​(π+α)−4

Now let, $\displaystyle {I}_{1}={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(\frac{{x}^{2}\mathrm{cos}x}{1+{\pi }^{x}})dx.......(i)$I1​=∫2−π​2π​​(1+πxx2cosx​)dx.......(i)

Now using the property $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx$∫ab​f(x)dx=∫ab​f(a+b−x)dx we get,

$$\Rightarrow {I}_{1}={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(\frac{{x}^{2}\mathrm{cos}x}{1+{\pi }^{-x}})dx........(ii)$$⇒I1​=∫2−π​2π​​(1+π−xx2cosx​)dx........(ii)

Now adding above equations we get,

$\displaystyle \Rightarrow 2{I}_{1}=2{\int }_{0}^{\frac{\pi }{2}}({x}^{2}\mathrm{cos}x)dx$⇒2I1​=2∫02π​​(x2cosx)dx $\displaystyle {\text{as}{\int }_{-a}^{a}f(x)dx=2{\int }_{0}^{a}f(x)dx,\text{if}f(x)\text{is even}}$as∫−aa​f(x)dx=2∫0a​f(x)dx,iff(x)is even

$$\Rightarrow {I}_{1}={[{x}^{2}(\mathrm{sin}x)]}_{0}^{\frac{\pi }{2}}-{\int }_{0}^{\frac{\pi }{2}}2x\cdot \mathrm{sin}xdx$$⇒I1​=[x2(sinx)]02π​​−∫02π​​2x⋅sinxdx

$$\Rightarrow {I}_{1}=\frac{{\pi }^{2}}{4}-[{[-2x\mathrm{cos}x]}_{0}^{\frac{\pi }{2}}+{\int }_{0}^{\frac{\pi }{2}}2\mathrm{cos}xdx]$$⇒I1​=4π2​−[[−2xcosx]02π​​+∫02π​​2cosxdx]

$$\Rightarrow {I}_{1}=\frac{{\pi }^{2}}{4}-2$$⇒I1​=4π2​−2

Similarly solving, $\displaystyle {I}_{2}={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(\frac{1+{\mathrm{sin}}^{2}x}{1+{e}^{{\mathrm{sin}}^{2023}x}})dx$I2​=∫2−π​2π​​(1+esin2023x1+sin2x​)dx

$$\Rightarrow {I}_{2}={\int }_{0}^{\frac{\pi }{2}}(1+{\mathrm{sin}}^{2}x)dx$$⇒I2​=∫02π​​(1+sin2x)dx

$$\Rightarrow {I}_{2}={\int }_{0}^{\frac{\pi }{2}}(1+\frac{1-\mathrm{cos}2x}{2})dx$$⇒I2​=∫02π​​(1+21−cos2x​)dx

$$\Rightarrow {I}_{2}=\frac{1}{2}{\int }_{0}^{\frac{\pi }{2}}(3-\mathrm{cos}2x)dx$$⇒I2​=21​∫02π​​(3−cos2x)dx

$$\Rightarrow {I}_{2}=\frac{1}{2}{[3x-\frac{\mathrm{sin}2x}{2}]}_{0}^{\frac{\pi }{2}}$$⇒I2​=21​[3x−2sin2x​]02π​​

$$\Rightarrow {I}_{2}=\frac{3\pi }{4}$$⇒I2​=43π​

$$\Rightarrow {I}_{1}+{I}_{2}=\frac{{\pi }^{2}}{4}-2+\frac{3\pi }{4}$$⇒I1​+I2​=4π2​−2+43π​

$$\Rightarrow \frac{\pi }{4}(\pi +3)-2=\frac{\pi }{4}(\pi +\alpha )-2$$⇒4π​(π+3)−2=4π​(π+α)−2

$$\Rightarrow \alpha =3$$⇒α=3