JEE Main 2024 — Mathematics Calculus

Given: $\displaystyle I=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{8\sqrt{2}\mathrm{cos}x}{(1+{e}^{\mathrm{sin}x})(1+{\mathrm{sin}}^{4}x)}dx...(i)$I=∫−2π​2π​​(1+esinx)(1+sin4x)82​cosx​dx...(i)

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

$$\Rightarrow I=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{8\sqrt{2}\mathrm{cos}x({e}^{\mathrm{sin}x})}{(1+{e}^{\mathrm{sin}x})(1+{\mathrm{sin}}^{4}x)}dx...(ii)$$⇒I=∫−2π​2π​​(1+esinx)(1+sin4x)82​cosx(esinx)​dx...(ii)

Adding $(i)\text{and}(ii)$(i)and(ii)

$$\Rightarrow 2I=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{8\sqrt{2}\mathrm{cos}x}{1+{\mathrm{sin}}^{4}x}dx$$⇒2I=∫−2π​2π​​1+sin4x82​cosx​dx

$$\Rightarrow I=\int _{0}^{\frac{\pi }{2}}\frac{8\sqrt{2}\mathrm{cos}x}{1+{\mathrm{sin}}^{4}x}dx$$⇒I=∫02π​​1+sin4x82​cosx​dx

Putting, $\mathrm{sin}x=t$sinx=t

$$\Rightarrow I=\int _{0}^{1}\frac{8\sqrt{2}}{1+{t}^{4}}dx$$⇒I=∫01​1+t482​​dx

$$\Rightarrow I=4\sqrt{2}\int _{0}^{1}(\frac{1+\frac{1}{{t}^{2}}}{{t}^{2}+\frac{1}{{t}^{2}}}-\frac{1-\frac{1}{{t}^{2}}}{{t}^{2}+\frac{1}{{t}^{2}}})dt$$⇒I=42​∫01​(t2+t21​1+t21​​−t2+t21​1−t21​​)dt

$$\Rightarrow I=4\sqrt{2}\int _{0}^{1}(\frac{1+\frac{1}{{t}^{2}}}{{(t-\frac{1}{t})}^{2}+2}-\frac{(1-\frac{1}{{t}^{2}})}{{(t+\frac{1}{t})}^{2}-2})dt$$⇒I=42​∫01​((t−t1​)2+21+t21​​−(t+t1​)2−2(1−t21​)​)dt

Let $t-\frac{1}{t}=z$t−t1​=z and $t+\frac{1}{t}=k$t+t1​=k

$$\Rightarrow I=4\sqrt{2}[\int _{-\infty }^{0}\frac{dz}{{z}^{2}+2}-\int _{\infty }^{2}\frac{dk}{{k}^{2}-2}]$$⇒I=42​[∫−∞0​z2+2dz​−∫∞2​k2−2dk​]

$$\Rightarrow I=4\sqrt{2}{[\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{z}{\sqrt{2}}]}_{-\infty }^{0}-{[\frac{1}{2\sqrt{2}}\mathrm{ln}(\frac{k-\sqrt{2}}{k+\sqrt{2}})]}_{\infty }^{2}$$⇒I=42​[2​1​tan−12​z​]−∞0​−[22​1​ln(k+2​k−2​​)]∞2​

$$\Rightarrow I=4\sqrt{2}[\frac{\pi }{2\sqrt{2}}-\frac{1}{2\sqrt{2}}[\mathrm{ln}\frac{2-\sqrt{2}}{2+\sqrt{2}}]]$$⇒I=42​[22​π​−22​1​[ln2+2​2−2​​]]

$$\Rightarrow I=2\pi +2\mathrm{log}(3+2\sqrt{2})$$⇒I=2π+2log(3+22​)

Comparing this value with $[\alpha \pi +\beta {\mathrm{log}}_{e}(3+2\sqrt{2})]$[απ+βloge​(3+22​)]

$\Rightarrow \alpha =2$⇒α=2 and $\beta =2$β=2

$$\Rightarrow {\alpha }^{2}+{\beta }^{2}=8$$⇒α2+β2=8