What is #int xtan^3 (5x)* sec^6 (5x)dx#?

1 Answer

#int x* tan^3 (5x)* sec^6 (5x) dx#

#=x/40 tan^8(5x)-1/1400 tan^7(5x)+x/15 tan^6(5x)-1/600 tan^5(5x)+x/20 tan^4 (5x)-1/1800 tan^3(5x)+1/600 tan(5x)-x/120+C#

Explanation:

Use integration by parts formula #int u dv=uv-int v du#

from the given #int x* tan^3 (5x)* sec^6 (5x) dx#

Let #u = x#
#dv = tan^3(5x)# #sec^6(5x) dx#

expand using Trigonometric identities

#dv=tan^3(5x)# # sec^4(5x)# # sec^2(5x) dx#
#dv=tan^3(5x)(tan^2(5x)+1)^2* sec^2(5x) dx#
#dv= tan^3(5x)[tan^4(5x)+2 tan^2(5x) +1] sec^2(5x) dx#
#dv=[tan^7(5x)+2 tan^5(5x) + tan ^3(5x)] sec^2(5x) dx#

after integration

#v=1/40 tan^8(5x) +1/15 tan ^6(5x) +1/20 tan^4(5x)#

#du = dx#

Use now the formula #int u dv=uv-int v du#