Question

Q. Find the reduction formula of ʃtan6xdx

Answer 1

Answer on Question #53132 – Math – Integral Calculus

Find the reduction formula of Jtan6xdx

Solution

$$I_n = \int \tan^n x \, dx = \int \tan^{n-2} x \tan^2 x \, dx = \int \tan^{n-2} x (\sec^2 x - 1) \, dx$$ $$= - \int \tan^{n-2} x \, dx + \int \tan^{n-2} x \sec^2 x \, dx.$$ $$\sec^2 x = (\tan x)'.$$ $$\int \tan^{n-2} x \sec^2 x \, dx = \int \tan^{n-2} x \, d(\tan x) = \frac{\tan^{n-1} x}{n - 1}.$$

The reduction formula of $I_n$:

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

The reduction formula of $I_6$:

$$I_6 = - I_4 + \frac{\tan^5 x}{5}.$$ $$I_4 = - I_2 + \frac{\tan^3 x}{3}.$$ $$I_2 = - I_0 + \tan x = - x + \tan x + C.$$

Thus

$$\int \tan^6 x \, dx = - \left( -(-x + \tan x) + \frac{\tan^3 x}{3} \right) + \frac{\tan^5 x}{5} + C = -x + \frac{\tan^5 x}{5} - \frac{\tan^3 x}{3} + \tan x + C,$$

where $C$ is an arbitrary real constant.

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