Question

If $I_n=\int {\tan }^{n}xdx$ then
$I_0+I_1+2(I_2+I_3+....+I_8)+I_9+I_{10}$ equals

• A. $1+\sum _{n=1}^8\frac{{\tan }^{n}x}{n}$
• B. $\sum _{n=1}^9\frac{{\tan }^{n}x}{n}$
• C. $\sum _{n=2}^{\infty }\frac{{\tan }^{n}x}{n+1}$
• D. $\sum _{n=1}^9\frac{{\tan }^{n}x}{n+1}$

Answer 1

The correct option is B $\sum _{n=1}^9\frac{{\tan }^{n}x}{n}$

$I_n=\int {\tan }^{n}xdx=\int {\tan }^{n-2}x({\sec }^{2}x-1)dx$
$\therefore I_n=\int {\tan }^{n-2}x{\sec }^{2}xdx-\int {\tan }^{n-2}xdx$
$\therefore I_{n-2}+I_n=\int {\tan }^{n-2}x{\sec }^{2}xdx=\frac{{\tan }^{n-1}x}{n-1}(\forall n\ge 2)$
Putting $n=2,3,4,...10$&adding we get
$\therefore I_0+I_1+2(I_2+I_3+I_4+...+I_8)+I_9+I_{10}=\sum _{n=1}^9\frac{{\tan }^{n}x}{n}$