$\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos \left(t^{\frac{1}{3}}\right) d t\right)$ is equal to [JEE Main 2024, 29 Jan (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: b) $\frac{3 \pi^{2}}{8}$. $L=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos \left(t^{\frac{1}{3}}\right) d t\right)$ Applying L-Hospital's Rule $\Rightarrow L=\lim _{x\to \frac{\pi }{2}}\frac{−\cos ⁡(x)3{x}^{2}}{2\left(x−\frac{\pi }{2}\right)}=\lim _{x\to \frac{\pi }{2}}\frac{\sin ⁡\left(x−\frac{\pi }{2}\right)3{x}^{2}}{2\left(x−\frac{\pi }{2}\right)}$ $\Rightarrow L=\frac{3{\pi }^{2}}{8}$

\(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\…

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