# If $I = \int_{0}^{\pi/2} \frac{dx}{\sqrt{1 + sin^3 x}}$ , then

Q: If $\displaystyle I = \int_{0}^{\pi/2} \frac{dx}{\sqrt{1 + sin^3 x}}$ , then

(A) 0 < I < 1

(B) $I > \frac{\pi}{2\sqrt{2}}$

(C) I < √2 π

(D) I > 2 π

Ans: (B) , (C)

Sol: Since x ∈ [0 , π/2 ]

⇒ 1 ≤ 1 + sin3 x  ≤  2

$\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{1 + sin^3 x}} \le 1$

$\int_{0}^{\pi/2} \frac{1}{\sqrt{2}} dx \le \int_{0}^{\pi/2} \frac{1}{\sqrt{1 + sin^3 x}} dx \le \int_{0}^{\pi/2} 1 dx$

$\frac{\pi}{2 \sqrt{2}} \le I \le \frac{\pi}{2}$

Hence (B) and (C) are the correct answers.