Results 1 to 5 of 5

Math Help - Integration by Trig Substitution

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    76

    Integration by Trig Substitution

    I am trying to integrate the following indefinite integral. 1/(x^2+2x+2)^2. I have this equal to 1/((x+1)^2+1)^2. The sides of the triangle are thus: hypotenuse = (x+1)^2+1,side A = x+1,side B = 1. Therefore with trig substitution I get the following equation to integrate: secant^2 theta/secant^2 theta. However I should get secant^2 theta/secant^4 theta. Can somebody point out where I have screwed up?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by p75213 View Post
    I am trying to integrate the following indefinite integral. 1/(x^2+2x+2)^2. I have this equal to 1/((x+1)^2+1)^2. The sides of the triangle are thus: hypotenuse = (x+1)^2+1,side A = x+1,side B = 1. Therefore with trig substitution I get the following equation to integrate: secant^2 theta/secant^2 theta. However I should get secant^2 theta/secant^4 theta. Can somebody point out where I have screwed up?
    What is your substitution ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    76
    x = tan theta - 1 -> secant^2 theta dtheta = dx
    secant theta = (x+1)^2 + 1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by p75213 View Post
    x = tan theta - 1 -> secant^2 theta dtheta = dx
    secant theta = (x+1)^2 + 1
    Note that: there is a square on the whole denominator in the original integral ..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2009
    Posts
    76
    Yes. The legs of the triangle are wrong. I guess I am asking how to arrive at the correct length of each leg. This one has me stumped. I'm sure it's not difficult but for some reason I just can't see it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration with Trig. Substitution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 19th 2010, 05:04 PM
  2. Integration by Trig substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 11th 2010, 03:32 AM
  3. Integration by trig substitution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 8th 2009, 12:52 AM
  4. integration with trig substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 18th 2009, 07:51 PM
  5. Integration by trig substitution
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 4th 2008, 01:19 PM

Search Tags


/mathhelpforum @mathhelpforum