Question

Obtain reduction formula for $I_n=\int {\tan }^{n}xdx,n$ being a positive integer $n\ge 2$ and deduce the value of $\int {\tan }^{6}xdx.$

Answer 1

Let $I_n=\int {\tan }^{n}xdx$

$\Rightarrow I_n=\int {\tan }^{2}x\cdot {\tan }^{n-2}xdx$

$\Rightarrow I_n=\int ({\sec }^{2}x-1){\tan }^{n-2}xdx$

$\Rightarrow I_n=\int {\sec }^{2}x{\tan }^{n-2}xdx-\int {\tan }^{n-2}xdx$

Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$\therefore I_n=\int t^{n-2}dt-\int {\tan }^{n-2}xdx$

$\Rightarrow I_n=\frac{t^{n-1}}{n-1}-\int {\tan }^{n-2}xdx$

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

Therefore,

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

Now,

$\int {\tan }^{6}xdx$ $=\frac{{\tan }^{5}x}{5}-\int {\tan }^{4}xdx=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}-\int {\tan }^{2}xdx=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}-\int (1-{\sec }^{2}x)dx=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}-\int dx+\int {\sec }^{2}xdx=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}-x+\tan x+C$

Therefore,

$\int {\tan }^{6}xdx=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}-x+\tan x+C$