Question

If $I_n=\int ta{n}^{n}xdxand{I}_n=\frac{ta{n}^{n-1}x}{n-1}+\lambda I_{n-2}then\lambda$ is equal to

Answer 1

${\tan }^{n}x={\tan }^{n-2}x{\tan }^{2}x$

$={\tan }^{n-2}x[{\sec }^{2}x-1]$

$={\tan }^{n-2}x{\sec }^{2}x-{\tan }^{n-2}x$

$\therefore I=\int {\tan }^{n}xdx=\int {\tan }^{n-2}x{\sec }^{2}xdx-\int {\tan }^{n-2}xdx$

substitute $u=\tan x\to du={\sec }^{2}xdx$

$=\int u^{n-2}du-\int {\tan }^{n-2}xdx$

$=\frac{u^{n-1}}{n-1}-\int {\tan }^{n-2}xdx$

$\int {\tan }^{n}xdx=\frac{{\tan }^{n-1}x}{n-1}-\int {\tan }^{n-2}xdx$

$\therefore I=\frac{{\tan }^{n-1}x}{n-1}-I_{n-2}$

$\Rightarrow \lambda =-1$