Question #98ecf

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Answer 1 · 2017-11-26T09:35:16

The answer is #=1/12tan^6(2x)+1/8tan^8(2x)+1/10tan^10(2x)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

Perform this integral by substitution

Let #u=2x#, #=>#, #du=2xdx#

So,

#inttan^5(2x)sec^6(2x)dx=1/2inttan^5(u)sec^6(u)du#

But

#sec^2(u)=tan^2(u)+1#

Therefore,

#1/2inttan^5(u)sec^6(u)du=1/2intsec^2(u)tan^5(u)(1+tan^2(u))^2du#

Now,

Let #v=tan(u)#, #=>#, #dv=sec^2(u)du#

So,

#1/2intsec^2(u)tan^5(u)(1+tan^2(u))^2du=1/2intv^5(1+v^2)^2du#

#=1/2intv^5(1+2v^2+v^4)dv#

#=1/2int(v^5+2v^7+v^9)dv#

#=1/2(1/6v^6+2/8v^8+1/10v^10)#

#=1/12v^6+1/8v^8+1/20v^10#

#=1/12tan^6(u)+1/8tan^8(u)+1/10tan^10(u)#

#=1/12tan^6(2x)+1/8tan^8(2x)+1/10tan^10(2x)+C#