How do you find #int xe^x - sec(7x)tan(7x)#?

1 Answer
Jan 28, 2017

I found: #xe^x-e^x-1/7*1/cos(7x)+c#

Explanation:

Let us write it as:
#intxe^xdx-intsec(x)tan(x)dx=#
we can use Integration by Parts for the first and a transformation for the second:
#=xe^x-int1e^xdx-int1/cos(7x)*sin(7x)/cos(7x)dx=#
#=xe^x-e^x-intsin(7x)/cos^2(7x)dx=#

we can consider that: #d[cos(7x)]=-7sin(7x)dx# and use this into the second integral and write (substituting and dividing by #7#):
#=xe^x-e^x+1/7int(d[cos(7x)])/cos^2(7x)=#
now we can integrate the second integral using #cos(7x)# as if it was our variable of integration and using the usual rules of integration to get:
#=xe^x-e^x-1/7*1/cos(7x)+c#