Question #8ec3f

Question

219 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-16T18:24:21

# 1/99*tan^9x(9tan^2x+11)+C#, or,

#1/99*tan^9x(9sec^2x+2)+C#.

I hope, the Question is to find #inttan^8xsec^4x dx#.

Suppose that, #I=inttan^8xsec^4x dx#,

#=inttan^8xsec^2xsec^2x dx#

#=inttan^8x(tan^2x+1)sec^2x dx#,

#=int(tan^10 x+tan^8 x)sec^2xdx#.

Here, we subst., #tanx=t," so that, "sec^2xdx=dt#.

#:. I=int(t^10+t^8)dt#,

#=t^11/11+t^9/9#,

#1/99*t^9(9t^2+11)#,

#=1/99*tan^9x(9tan^2x+11)+C, or, #

#=1/99*tan^9x{9(tan^2x+1)+2}#,

#=1/99*tan^9x(9sec^2x+2)+C#.