Question

Let $I_1={\int }_0^{\frac{\pi }{2}}(\cos x{)}^{\sqrt{13}+1}dx$ and $I_2={\int }_0^{\frac{\pi }{2}}(\cos x{)}^{\sqrt{13}-1}dx$, then the value of $[\frac{I_2}{I_1}]$ is (where $[\cdot ]$ is the greatest integer function)

• A. 2
• B. 1
• C. A prime number
• D. Less than 3

Answer 1

Answer: (B), (D)

The correct options are
B 1
D Less than 3

$I_1={\int }_0^{\frac{\pi }{2}}(\cos x{)}^{\sqrt{13}}.\cos xdx=[(\cos x{)}^{\sqrt{13}}\sin x{]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\sqrt{13}(\cos x{)}^{\sqrt{13}-1}.(-\sin x)\left(\sin x\right)dx=\sqrt{13}{\int }_0^{\frac{\pi }{2}}\left(\cos x{)}^{\sqrt{13}-1}\right(1-co{s}^{2}x)dx{I}_1=\sqrt{13}(I_2-I_1)\frac{I_2}{I_1}=1+\frac{1}{\sqrt{13}}\Rightarrow [\frac{I_2}{I_1}]=1$