Question

Let $I_n=\int {\tan }^{n}xdx,(n>1)$. If $I_4+I_6=a{\tan }^{5}x+b{x}^5+C$, where $C$ is a constant of integration, then the ordered pair $(a,b)$ is equal to

• A. $(-\frac{1}{5},1)$
• B. $\left(\frac{1}{5},0\right)$
• C. $\left(\frac{1}{5},-1\right)$
• D. $(-\frac{1}{5},0)$

Answer 1

Answer: (B)

$\left(\frac{1}{5},0\right)$

Given that: $\int {\tan }^{n}xdx$

Using induction formula for $\tan$ we get:

$I_n=\frac{{\tan }^{(n-1)}x}{(n-1)}-I_{(n-2)}+c$

$I_n+I_{(n-2)}=\frac{ta{n}^{n-1}x}{(n-1)}+c$

Hence $I_6+I_4=\frac{{\tan }^{6-1}x}{6-1}+c$

$=\frac{{\tan }^{5}x}{5}+0x+c$

$\therefore a=\frac{1}{5},b=0$