Question

If $I_n=\underset{0}{\overset{\pi /4}{\int }}ta{n}^{n}xdx$, where $n$ is a positive integer, then $I_{10}+I_8$ is

• A. $\frac{1}{9}$
• B. $\frac{1}{8}$
• C. $\frac{1}{7}$
• D. $9$

Answer 1

Answer: (A)

Given integral is,
$I_n=\underset{0}{\overset{\pi /1}{\int }}{\tan }^{n}xdx$
$I_{\vec{n}}=\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n-2}x\cdot {\tan }^{2}xdx$
$=\underset{0}{\overset{\pi 4}{\int }}{\tan }^{n-2}x\cdot \left({\sec }^{2}x-1\right)dx$
$I_n=\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n-2}x\cdot {\sec }^{2}xdx$
$-\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n-2}xdx$
$I_n=\underset{0}{\overset{1}{\int }}{t}^{n-2}dt-\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n-2}xdx$
$I_n={\left[\frac{t^{n-1}}{n-1}\right]}_0^1-I_{n-2}$
$$\begin{gathered} \left\{\text{ put }\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt\right\} \end{gathered}$$
$I_n+I_{n-2}=\frac{1}{n-1}$
Put $n=10$, we get
$I_{10}+I_8=\frac{1}{9}$