using the reduction formula of secx^nsecxn
intsecx^ndx∫secxndx=intsecx^(n-2)*secx^2dx∫secxn−2⋅secx2dx
integrate intsecx^(n-2)*secx^2dx∫secxn−2⋅secx2dx by parts (uv- intvduuv−∫vdu)
where u=secx^(n-2)u=secxn−2 and dv=secx^2dv=secx2
therefore intsecx^(n-2)*secx^2dx∫secxn−2⋅secx2dx
=tanx*secx^(n-2)-(n-2)inttan^2*secx^(n-2) dxtanx⋅secxn−2−(n−2)∫tan2⋅secxn−2dx
=tanx*secx^(n-2)-(n-2)int(secx^2-1)*secx^(n-2)dxtanx⋅secxn−2−(n−2)∫(secx2−1)⋅secxn−2dx
=tanx*secx^(n-2)-(n-2)intsecx^n*secx^(n-2)dxtanx⋅secxn−2−(n−2)∫secxn⋅secxn−2dx
if I_"n"=intsecx^ndxIn=∫secxndx
therefore
I_"n"=tanx*secx^(n-2)-(n-2)intsecx^n-secx^(n-2)dxIn=tanx⋅secxn−2−(n−2)∫secxn−secxn−2dx
I_"n"=tanx*secx^(n-2)-(n-2)I_n+(n-2)I_"n-2"In=tanx⋅secxn−2−(n−2)In+(n−2)In-2
(n-1)I_"n"=tanx*secx^(n-2)+(n-2)I_"n-2"(n−1)In=tanx⋅secxn−2+(n−2)In-2
I_"n"=1/(n-1)*(tanx*secx^(n-2))+(n-2)/(n-1)I_"n-2"In=1n−1⋅(tanx⋅secxn−2)+n−2n−1In-2
Where I_"n"=intsecx^ndxIn=∫secxndx and I_"n-2"In-2=intsecx^(n-2)dx∫secxn−2dx
Substitute every n with 4 in I_"n"=1/(n-1)*(tanx*secx^(n-2))+(n-2)/(n-1)I_"n-2"In=1n−1⋅(tanx⋅secxn−2)+n−2n−1In-2
Therefore
intsecx^4dx=1/3 secx^2*tanx-1/2intsecx^2dx∫secx4dx=13secx2⋅tanx−12∫secx2dx
=1/3 secx^2*tanx-1/2tanx13secx2⋅tanx−12tanx
This method can be used to integrate sec x to the power of anything as long as the power is bigger than 1
