JEE Main 2022 — Mathematics Calculus

Let, $\displaystyle I=\frac{24}{\pi }{\int }_{0}^{\sqrt{2}}\frac{2-{x}^{2}}{(2+{x}^{2})(\sqrt{4+{x}^{4}})}dx$I=π24​∫02​​(2+x2)(4+x4​)2−x2​dx

$$=\frac{24}{\pi }{\int }_{0}^{\sqrt{2}}\frac{2-{x}^{2}}{{x}^{2}(\frac{2}{x}+x)\sqrt{(\frac{4}{{x}^{2}}+{x}^{2})}}dx$$=π24​∫02​​x2(x2​+x)(x24​+x2)​2−x2​dx

$$=\frac{24}{\pi }{\int }_{0}^{\sqrt{2}}\frac{\frac{2}{{x}^{2}}-1}{(\frac{2}{x}+x)\sqrt{{(\frac{2}{x}+x)}^{2}-4}}dx$$=π24​∫02​​(x2​+x)(x2​+x)2−4​x22​−1​dx

Now let $\frac{2}{x}+x=t,(-\frac{2}{{x}^{2}}+1)dx=dt$x2​+x=t,(−x22​+1)dx=dt

$$I=-\frac{24}{\pi }{\int }_{\infty }^{2\sqrt{2}}\frac{dt}{t\sqrt{{t}^{2}-4}}=-\frac{12}{\pi }{\int }_{\infty }^{2\sqrt{2}}\frac{2tdt}{{t}^{2}\sqrt{{t}^{2}-4}}$$I=−π24​∫∞22​​tt2−4​dt​=−π12​∫∞22​​t2t2−4​2tdt​

Let ${t}^{2}-4={z}^{2},2tdt=2zdz$t2−4=z2,2tdt=2zdz

$$I=-\frac{12}{\pi }{\int }_{\infty }^{2}\frac{2zdz}{z({z}^{2}+4)}=-\frac{24}{\pi }{\int }_{\infty }^{2}\frac{dz}{({z}^{2}+4)}=-\frac{24}{\pi }{(\frac{1}{2}{\mathrm{tan}}^{-1}\frac{z}{2})}_{\infty }^{2}$$I=−π12​∫∞2​z(z2+4)2zdz​=−π24​∫∞2​(z2+4)dz​=−π24​(21​tan−12z​)∞2​

$$=-\frac{24}{\pi }(\frac{\pi }{8}-\frac{\pi }{4})=3$$=−π24​(8π​−4π​)=3