Question

If $I_n={\int }_0^{\pi /4}{\tan }^{n}xdx$, then

• A. $I_7+I_5=\frac{1}{6}$
• B. $I_{10}+I_8=\frac{1}{9}$
• C. $I_8-I_{12}=\frac{2}{99}$
• D. $I_{12}+2I_{10}+I_8=\frac{20}{99}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $I_{10}+I_8=\frac{1}{9}$
B $I_7+I_5=\frac{1}{6}$
C $I_8-I_{12}=\frac{2}{99}$
D $I_{12}+2I_{10}+I_8=\frac{20}{99}$

$I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}xdx$
Replacing $n\to n+1$
$I_{n+1}={\int }_0^{\frac{\pi }{4}}{\tan }^{n+1}xdx$ ...(1)
Replacing $n\to n-1$
$I_{n-1}={\int }_0^{\frac{\pi }{4}}{\tan }^{n-1}xdx$
$I_{n+1}+I_{n-1}={\int }_0^{\frac{\pi }{4}}\left({\tan }^{n+1}x+{\tan }^{n-1}x\right)dx=\frac{1}{n}$
$\therefore I_7+I_5=\frac{1}{6}$
$I_{10}+I_8=\frac{1}{9}$
$I_8-I_{12}=I_8+I_{10}-(I_{10}+I_{12})=\frac{1}{9}-\frac{1}{11}=\frac{2}{99}$