Integrating Trig Functions: Solving Trig Integration Homework

[nameVoid]

Homework Statement

I(tan^5x,x)=I(tan^2xtan^3x,x)=I((sec^2x-1)tan^3x,x)
=I(sec^2xtan^3x,x)-I(tan^3x,x)
u=tanx du=sec^2x
=I(u^3,u)-I(tan^3x,x)
=u^4/4-I(tan^3x,x)
=tan^4x/4-I(tan^2xtanx,x)
=tan^4x/4-I((sec^2x-1)tanx,x)
=tan^4x/4-I(tanxsec^2x-tanx)
=tan^4x/4-I(tanxsec^2x,x)-I(tanx,x)
y=tanx, dy=sec^2x
=tan^4x/4-I(u,u)-I(tanx,x)
=tan^4x/4-u^2/2-ln|secx|+c
tan^4x/4-tan^2x/2-ln|secx|+c

Homework Equations

The Attempt at a Solution

 

[Bohrok]

You have a sign error between lines 8 and 9

=tan^4x/4-I(tanxsec^2x-tanx)
=tan^4x/4-I(tanxsec^2x,x)-I(tanx,x)

 

[nameVoid]

I(tan^5x,x)=I(tan^2xtan^3x,x)=I((sec^2x-1)tan^3x,x)
=I(sec^2xtan^3x,x)-I(tan^3x,x)
u=tanx du=sec^2x
=I(u^3,u)-I(tan^3x,x)
=u^4/4-I(tan^3x,x)
=tan^4x/4-I(tan^2xtanx,x)
=tan^4x/4-I((sec^2x-1)tanx,x)
=tan^4x/4-I(tanxsec^2x-tanx)
=tan^4x/4-I(tanxsec^2x,x)+I(tanx,x)
y=tanx, dy=sec^2x
=tan^4x/4-I(y,y)+I(tanx,x)
=tan^4x/4-y^2/2+ln|secx|+c
tan^4x/4-tan^2x/2+ln|secx|+c

my texts solution is 1/4sec^4x-sec^2x+ln|secx|+c

 

[Cyosis]

To obtain your text's solution you can simply do what you've been doing all along. That is use the identity $\tan^2x=\sec^2x-1$.