JEE Main 2025 — Mathematics Calculus

I.F. $\displaystyle \mathrm{e}^{-\frac{1}{2} \int \frac{2 \mathrm{x}}{1-\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{-\frac{1}{2} \ln \left(1-\mathrm{x}^2\right)}=\sqrt{1-\mathrm{x}^2}$e−21​∫1−x22x​dx=e−21​ln(1−x2)=1−x2​ $\displaystyle y \times \sqrt{1-x^2}=\int\left(x^6+4 x\right) d x=\frac{x^7}{7}+2 x^2+c$y×1−x2​=∫(x6+4x)dx=7x7​+2x2+c Given $y(0)=0 \Rightarrow c=0$y(0)=0⇒c=0 $y=\frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}}$y=1−x2​7x7​+2x2​ Now, $\displaystyle 6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{x^7}{7}+2 \mathrm{x}^2}{\sqrt{1-\mathrm{x}^2}} \mathrm{dx}=6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2 \mathrm{x}^2}{\sqrt{1-\mathrm{x}^2}} \mathrm{dx}$6∫−21​21​​1−x2​7x7​+2x2​dx=6∫−21​21​​1−x2​2x2​dx $\displaystyle \begin{aligned} & =24 \int_0^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} d x \\ & \text { Put } x=\sin \theta \\ & d x=\cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}} \frac{\sin ^2 \theta}{\cos \theta} \cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta=12\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{6}} \\ & =12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \\ & =2 \pi-3 \sqrt{3} \\ & \alpha^2=(3 \sqrt{3})^2=27\end{aligned}$​=24∫021​​1−x2​x2​dx Put x=sinθdx=cosθdθ=24∫06π​​cosθsin2θ​cosθdθ=24∫06π​​(21−cos2θ​)dθ=12[θ−2sin2θ​]06π​​=12(6π​−43​​)=2π−33​α2=(33​)2=27​