• Support PF! Buy your school textbooks, materials and every day products Here!

Trig integration

  • Thread starter nameVoid
  • Start date
  • #1
242
0

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

 

Answers and Replies

  • #2
867
0
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)
 
Last edited:
  • #3
242
0
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
 
  • #4
Cyosis
Homework Helper
1,495
0
To obtain your text's solution you can simply do what you've been doing all along. That is use the identity [itex]\tan^2x=\sec^2x-1[/itex].
 
Last edited:

Related Threads for: Trig integration

  • Last Post
Replies
4
Views
887
  • Last Post
Replies
2
Views
746
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
Top