Question #73da1

1 Answer
Jan 7, 2018

#inttan^5xsec^7xdx=sec^11x/11-(2sec^9x)/9+sec^7x/7+"C"#

Explanation:

Given: #inttan^5xsec^7xdx#
#--------------------#
Make a substitution:

Let #u=secx#

Thus, #du=tanxsecxdx#

#inttan^5xsec^7xdx#

#->inttan^4xsec^6x(tanxsecxdx)#
#--------------------#
Also: Convert #tan^2x# to secants using the identity:

#tan^2x=sec^2x-1#

So #tan^4x=(sec^2x-1)^2#
#--------------------#
Rewriting the integral...

#=int(sec^2x-1)^2sec^6x(tanxsecxdx)#

#=int(sec^4x-2sec^2x+1)sec^6x(tanxsecxdx)#

#=intsec^10x-2sec^8x+sec^6x(tanxsecxdx)#

Since #u=secx# and #du=tanxsecxdx#, the integral is now

#intu^10-2u^8+u^6du#

#=u^11/11-(2u^9)/9+u^7/7+"C"#

Lastly, subsitute back #u=secx#

#=sec^11x/11-(2sec^9x)/9+sec^7x/7+"C"#