JEE Main 2024 — Definite Integration Question with Solution

I=\int_\limits0^{\frac{\pi}{2}} \sin 2 x \cdot(\cos x)^{\frac{11}{2}}\left(1+(\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} d x

Put $$\cos x=t^2\Rightarrow \sin xdx=-2tdt$$

\begin{aligned} & \therefore \mathrm{I}=4 \int_\limits0^1 \mathrm{t}^2 \cdot \mathrm{t}^{11} \sqrt{\left(1+\mathrm{t}^5\right)}(\mathrm{t}) \mathrm{dt} \\ & \mathrm{I}=4 \int_\limits0^1 \mathrm{t}^{14} \sqrt{1+\mathrm{t}^5} \mathrm{dt} \end{aligned}

Put $$1+{\mathrm{t}}^5={\mathrm{k}}^2$$

$$\Rightarrow 5{\mathrm{t}}^4\mathrm{dt}=2\mathrm{k}\mathrm{dk}$$

\therefore \mathrm{I}=4 \cdot \int_\limits1^{\sqrt{2}}\left(\mathrm{k}^2-1\right)^2 \cdot \mathrm{k} \frac{2 \mathrm{k}}{5} \mathrm{dk}

\mathrm{I}=\frac{8}{5} \int_\limits1^{\sqrt{2}} \mathrm{k}^6-2 \mathrm{k}^4+\mathrm{k}^2 \mathrm{dk}

$$\mathrm{I}=\frac{8}{5}{\left[\frac{{\mathrm{k}}^7}{7}-\frac{2{\mathrm{k}}^5}{5}+\frac{{\mathrm{k}}^3}{3}\right]}_1^{\sqrt{2}}$$

$$\mathrm{I}=\frac{8}{5}\left[\frac{8\sqrt{2}}{7}-\frac{8\sqrt{2}}{5}+\frac{2\sqrt{2}}{3}-\frac{1}{7}+\frac{2}{5}-\frac{1}{3}\right]$$

$$\begin{array}{rl}& \mathrm{I}=\frac{8}{5}\left[\frac{22\sqrt{2}}{105}-\frac{8}{105}\right]\\ & \therefore 525\cdot \mathrm{I}=176\sqrt{2}-64\end{array}$$