JEE Main 2024 — Mathematics Calculus

Let, $\displaystyle I=\int _{0}^{\frac{\pi }{4}}\frac{xdx}{{\mathrm{sin}}^{4}(2x)+{\mathrm{cos}}^{4}(2x)}$I=∫04π​​sin4(2x)+cos4(2x)xdx​

Let, $2x=t$2x=t then $dx=\frac{1}{2}dt$dx=21​dt

$$\Rightarrow I=\frac{1}{4}\int _{0}^{\frac{\pi }{2}}\frac{tdt}{{\mathrm{sin}}^{4}t+{\mathrm{cos}}^{4}t}$$⇒I=41​∫02π​​sin4t+cos4ttdt​

$$\Rightarrow I=\frac{1}{4}\int _{0}^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}-t)dt}{{\mathrm{sin}}^{4}(\frac{\pi }{2}-t)+{\mathrm{cos}}^{4}(\frac{\pi }{2}-t)}$$⇒I=41​∫02π​​sin4(2π​−t)+cos4(2π​−t)(2π​−t)dt​

$$\Rightarrow I=\frac{1}{4}\int _{0}^{\frac{\pi }{2}}\frac{\frac{\pi }{2}dt}{{\mathrm{sin}}^{4}t+{\mathrm{cos}}^{4}t}-I$$⇒I=41​∫02π​​sin4t+cos4t2π​dt​−I

$$\Rightarrow 2I=\frac{\pi }{8}\int _{0}^{\frac{\pi }{2}}\frac{dt}{{\mathrm{sin}}^{4}t+{\mathrm{cos}}^{4}t}$$⇒2I=8π​∫02π​​sin4t+cos4tdt​

$$\Rightarrow 2I=\frac{\pi }{8}\int _{0}^{\frac{\pi }{2}}\frac{{\mathrm{sec}}^{4}tdt}{{\mathrm{tan}}^{4}t+1}$$⇒2I=8π​∫02π​​tan4t+1sec4tdt​

Let, $\mathrm{tan}t=y$tant=y then ${\mathrm{sec}}^{2}tdt=dy$sec2tdt=dy

$$\Rightarrow 2I=\frac{\pi }{8}\int _{0}^{\infty }\frac{(1+{y}^{2})dy}{1+{y}^{4}}$$⇒2I=8π​∫0∞​1+y4(1+y2)dy​

$$\Rightarrow I=\frac{\pi }{16}\int _{0}^{\infty }\frac{1+\frac{1}{{y}^{2}}}{{y}^{2}+\frac{1}{{y}^{2}}}dy$$⇒I=16π​∫0∞​y2+y21​1+y21​​dy

Putting, $y-\frac{1}{y}=p$y−y1​=p

$$\Rightarrow I=\frac{\pi }{16}\int _{-\infty }^{\infty }\frac{dp}{{p}^{2}+{(\sqrt{2})}^{2}}$$⇒I=16π​∫−∞∞​p2+(2​)2dp​

$$\Rightarrow I=\frac{\pi }{16\sqrt{2}}{[{\mathrm{tan}}^{-1}(\frac{p}{\sqrt{2}})]}_{-\infty }^{\infty }$$⇒I=162​π​[tan−1(2​p​)]−∞∞​

$$\Rightarrow I=\frac{{\pi }^{2}}{16\sqrt{2}}$$⇒I=162​π2​

$$\Rightarrow I=\frac{\sqrt{2}{\pi }^{2}}{32}$$⇒I=322​π2​