Answer to Question #155793 in Calculus for Yolande

Question

Find the integral of: sec^n (x)dx

Answer 1

$I_{n}=\int sec^nxdx --------(1)$

$I_{n}=\int sec^{n-2}x.sec^2xdx$

Integration by using parts where as $\int sec^{n-2}xdx$ is $I_{st}$ and $\int sec^{2}xdx$ is $II_{nd}$ part

$I_{n}=sec^{n-2}x\int sec^2xdx-\int[d/dx(sec^{n-2}).\int sec^2xdx]dx$

Since $\int sec^2xdx = tanx$ and where as $d/dx(sec^{n-2}x=(n-2)sec^{n-3}xsecxtanx$

$I_{n}=sec^{n-2}x.tanx-\int(n-2)sec^{n-3}x.secx.tanx.tanxdx$

$=sec^{n-2}x.tanx-(n-2)\int sec^{n-2}x.tan^2xdx$

$=sec^{n-2}x.tanx-(n-2)\int sec^{n-2}x(sec^2x-1)dx$

$=sec^{n-2}x.tanx-(n-2)[\int sec^{n}xdx-\int sec^{n-2}xdx$

$=sec^{n-2}x.tanx-(n-2)[\int sec^{n}xdx-\int sec^{n-2}xdx]$

$=sec^{n-2}x.tanx-(n-2)[I_{n}-I_{(n-2)}]$

From equation (1) $I_{n}=\int sec^nxdx$ therefore $I_{n-2}=\int sec^{n-2}xdx$

$I_{n}=sec^{n-2}x.tanx-(n-2)I_{n}+(n-2)I_{(n-2)}$

$I_{n}+(n-2)I_{n}=sec^{n-2}x.tanx+(n-2)I_{n-2}$

$(n-1)I_{n}=sec^{n-2}x.tanx+(n-2)I_{n-2}$ $I_{n}=[sec^{n-2}x.tanx/(n-1)]+[(n-2)I_{n-2}/(n-1)]$

This is the reduction formula of $\int sec^{n-2}xdx$

Answer: $I_{n}=[sec^{n-2}x.tanx/(n-1)]+[(n-2)I_{n-2}/(n-1)]$