If $I=\int\limits_0^{\pi / 2} \frac{d x}{\sqrt{1+\sin ^3 x}}$, then
Answer & explanation
Correct answer: option 3
Since $x \in\left[0, \frac{\pi}{2}\right] \Rightarrow 1 \leq 1+\sin ^3 x \leq 2$
$\Rightarrow \frac{1}{\sqrt{2}} \leq \frac{1}{\sqrt{1+\sin ^3 x}} \leq 1 \Rightarrow \int\limits_0^{\pi / 2} \frac{1}{\sqrt{2}} d x \leq \int\limits \frac{d x}{\sqrt{1+\sin ^3 x}} \leq \int\limits_0^{\pi / 2} d x$
$\Rightarrow \frac{\pi}{2 \sqrt{2}} \leq I \leq \frac{\pi}{2}$
Hence (3) is the correct answer.