Question

Find the reduction formula of ʃsec6xdx.

Answer 1

Answer on Question #53134 – Math – Integral Calculus

Find the reduction formula of $\int \sec^6(x) \, dx$.

Solution

$$\begin{array}{l} I_n = \int \sec^n x \, dx = \int \sec^{n-2} x \sec^2 x \, dx = \int \sec^{n-2} x \, d(\tan x) = \\ = \sec^{n-2} x \tan x + (n - 2) \int \sec^{n-2} x \tan^2 x \, dx = \\ = \sec^{n-2} x \tan x + (n - 2) \int \sec^{n-2} x (\sec^2 x - 1) \, dx = \\ = \sec^{n-2} x \tan x + (n - 2) I_{n-2} - (n - 2) I_n \rightarrow \\ \rightarrow I_n = \frac{1}{n - 1} \sec^{n-2} x \tan x + \frac{n - 2}{n - 1} I_{n-2} \end{array}$$

Thus

$$\begin{array}{l} I_6 = \frac{1}{5} \sec^4 x \tan x + \frac{4}{5} \left( \frac{1}{3} \sec^2 x \tan x + \frac{2}{3} I_2 \right) = \\ = \frac{1}{5} \sec^4 x \tan x + \frac{4}{15} \sec^2 x \tan x + \frac{8}{15} \tan x + c, \end{array}$$

where $c$ is an arbitrary real constant.

So

$$\int \sec^6 x \, dx = \frac{\tan x}{15} (\sec^4 x + 4 \sec^2 x + 8) + c,$$

where $c$ is an arbitrary real constant.

www.AssignmentExpert.com