JEE Main 2024 — Mathematics Calculus

Let,

$$I=\int _{0}^{\frac{\pi }{2}}\mathrm{sin}2x{\mathrm{cos}}^{\frac{11}{2}}x{(1+{\mathrm{cos}}^{\frac{5}{2}}x)}^{\frac{1}{2}}dx$$I=∫02π​​sin2xcos211​x(1+cos25​x)21​dx

Let $\mathrm{cos}x={t}^{2}\Rightarrow \mathrm{sin}xdx=-2tdt$cosx=t2⇒sinxdx=−2tdt

$$\therefore I=-4\int _{1}^{0}{t}^{2}\cdot {t}^{11}\sqrt{(1+{t}^{5})}(t)dt$$∴I=−4∫10​t2⋅t11(1+t5)​(t)dt

$$\Rightarrow I=4\int _{0}^{1}{t}^{14}\sqrt{(1+{t}^{5})}dt$$⇒I=4∫01​t14(1+t5)​dt

Now, let $1+{t}^{5}={k}^{2}$1+t5=k2

$$\Rightarrow 5{t}^{4}dt=2kdk$$⇒5t4dt=2kdk

$$\therefore I=4\cdot \int _{1}^{\sqrt{2}}{({k}^{2}-1)}^{2}\cdot k\cdot \frac{2k}{5}dk$$∴I=4⋅∫12​​(k2−1)2⋅k⋅52k​dk

$$\Rightarrow I=\frac{8}{5}\int _{1}^{\sqrt{2}}{k}^{6}-2{k}^{4}+{k}^{2}dk$$⇒I=58​∫12​​k6−2k4+k2dk

$$\Rightarrow I=\frac{8}{5}{[\frac{{k}^{7}}{7}-\frac{2{k}^{5}}{5}+\frac{{k}^{3}}{3}]}_{1}^{\sqrt{2}}$$⇒I=58​[7k7​−52k5​+3k3​]12​​

$$\Rightarrow I=\frac{8}{5}[\frac{8\sqrt{2}}{7}-\frac{8\sqrt{2}}{5}+\frac{2\sqrt{2}}{3}-\frac{1}{7}+\frac{2}{5}-\frac{1}{3}]$$⇒I=58​[782​​−582​​+322​​−71​+52​−31​]

$$\Rightarrow I=\frac{8}{5}[\frac{22\sqrt{2}}{105}-\frac{8}{105}]$$⇒I=58​[105222​​−1058​]

$$\Rightarrow 525\cdot I=176\sqrt{2}-64$$⇒525⋅I=1762​−64

Hence, on comparing we get, $n=176$n=176