Question

If $I_1={\int }_0^{\pi /2}{\sin }^{4}xdx$
$I_2={\int }_0^{\pi /2}{\cos }^{6}xdx$
$I_3={\int }_0^{\pi /2}{\sin }^{8}xdx$
$I_4={\int }_0^{\pi /2}{\cos }^{2}xdx$

then the increasing order of $I_1,I_2,I_3,I_4$ is?

• A. $I_4,I_2,I_3,I_1$
• B. $I_3,I_2,I_1,I_4$
• C. $I_4,I_3,I_2,I_1$
• D. $I_3,I_1,I_2,I_4$

Answer 1

The correct option is B $I_3,I_2,I_1,I_4$

$I_1={\int }_0^{\pi /2}sin{x}^{4}dx;I_2{\int }_0^{\pi /2}co{s}^{6}xdx;I_3={\int }_0^{\pi /2}sin{x}^{8dx}$
$I_4{\int }_0^{\pi /2}cos{x}^{2}dx$
Increasing order.
$I_1=\frac{3}{4}\times \frac{1}{2}\times \pi /2;I_2=\frac{5}{6}\times \frac{3}{4}\times \frac{1}{2}\times \pi /2$
$I_3=\frac{7}{8}\times \frac{5}{6}\times \frac{3}{4}\times \frac{1}{2}\times \pi /2$
$I_4=\frac{1}{2}\times \pi /2$So, increasing order
$I_3,I_2,I_1,I_4$