Question

Let $I_n={\int }_0^{\pi /4}{\tan }^{n}xdx,nϵN.$ Then

• A. $I_1=I_3+2I_5$
• B. $I_n+I_{n-2}=\frac{1}{n}$
• C. $I_n+I_{n-2}=\frac{1}{n-1}$
• D. none of these

Answer 1

Answer: (A), (C)

The correct options are
A $I_1=I_3+2I_5$
C $I_n+I_{n-2}=\frac{1}{n-1}$

$I_n={\int }_0^{\pi /4}{\tan }^{n}xdx$
We have reduction formula for
$\int {\tan }^{n}xdx=\frac{1}{n-1}{\tan }^{n-1}x-\int {\tan }^{n-2}xdx$
So, $I_n=\frac{1}{n-1}{\left[{\tan }^{n-1}x\right]}_0^{\pi /4}-{\int }_0^{\pi /4}{\tan }^{n-2}xdx$
$\Rightarrow I_n=\frac{1}{n-1}-I_{n-2}$
$\Rightarrow I_n+I_{n-2}=\frac{1}{n-1}$
$I_5+I_3=\frac{1}{4}$
$I_3+I_1=\frac{1}{2}$
$\Rightarrow I_1=I_3+2I_5$
Hence, options 'A' and 'C' are correct.