If I=∫π20dx√1+sin3x, then
0<I<1
I>π2√2
I<√2π
I>2π
Since, x ϵ [0,π2]⇒ 1≤1+sin3x≤2⇒ 1√2≤1√1+sin3x≤1⇒ ∫π201√2dx≤∫π20dx√1+sin3x≤∫π20dx⇒π2√2≤I≤π2.