Results 1 to 8 of 8

Math Help - [SOLVED] 4 Integrals using substitution

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    [SOLVED] 4 Integrals using substitution

    I am stuck on 4 problems involving a substitution to be solved.
    It only says it must be solved using a substitution and no other method of solving integrals.
    1) \int \frac{e^x dx}{e^{2x}+2e^x+1}. I've trying a u-sub : u(x)=e^x, also u(x)=e^{\frac{x}{2}} and u(x)=e^{2x}+2e^x+1 but in each case I failed to solve it.
    2) \int ln(\cos(x))tan(x)dx. I don't really know what sub to make here. Maybe u=\cos (x)?
    3) \int \frac{dx}{e^x+e^{-x}}. Making the sub u(x)=e^x, I got \int \frac{dx}{e^x+e^{-x}}=\int \frac{du}{u^2-u} but I don't know how to finish it!
    4) \int \frac{dx}{1+e^x}. I took u(x)=1+e^x and I got an answer very similar to the 3) : \int \frac{du}{u^2+u} but don't know how to finish it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    Quote Originally Posted by arbolis View Post
    I am stuck on 4 problems involving a substitution to be solved.
    It only says it must be solved using a substitution and no other method of solving integrals.
    1) \int \frac{e^x dx}{e^{2x}+2e^x+1}. I've trying a u-sub : u(x)=e^x, also u(x)=e^{\frac{x}{2}} and u(x)=e^{2x}+2e^x+1 but in each case I failed to solve it.
    2) \int ln(\cos(x))tan(x)dx. I don't really know what sub to make here. Maybe u=\cos (x)?
    3) \int \frac{dx}{e^x+e^{-x}}. Making the sub u(x)=e^x, I got \int \frac{dx}{e^x+e^{-x}}=\int \frac{du}{u^2-u} but I don't know how to finish it!
    4) \int \frac{dx}{1+e^x}. I took u(x)=1+e^x and I got an answer very similar to the 3) : \int \frac{du}{u^2+u} but don't know how to finish it.

    1) your first sub is correct the integral becomes \int \frac{du}{u^2+2u+1} = \int \frac{du}{(u+1)^2}  which is friarly easy.

    2) observe \frac{d}{dx} ( ln(\cos(x))) = - \tan x

    3) partial fractions \frac{1}{u(u-1)} = \frac{1}{u-1} - \frac{1}{u}


    Bobak
    Last edited by bobak; July 6th 2008 at 06:47 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    follow bobak for the other two

    Quote Originally Posted by arbolis View Post
    3) \int \frac{dx}{e^x+e^{-x}}. Making the sub u(x)=e^x, I got \int \frac{dx}{e^x+e^{-x}}=\int \frac{du}{u^2-u} but I don't know how to finish it!
    actually, a substitution of u = e^x yields \int \frac 1{u^2 + 1}~du. which you should know

    4) \int \frac{dx}{1+e^x}. I took u(x)=1+e^x and I got an answer very similar to the 3) : \int \frac{du}{u^2+u} but don't know how to finish it.
    correct. by partial fractions (or otherwise) note that \frac 1{u^2 + u} = \frac 1{u(u + 1)} = \frac 1u - \frac 1{u + 1}. now continue
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thanks bobak!
    Yeah, I've reached
    \int \frac{dx}{u^2+2u+1} = \int \frac{dx}{(u+1)^2}
    but was stuck. I'll check a book for formulas then...
    About the rest, I'll try it tomorrow probably and will post any help I feel to ask.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    actually, a substitution of yields . which you should know
    I didn't reach this integral which is equal to arctan(u)! I will spend some time on it.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by arbolis View Post
    Thanks bobak!
    Yeah, I've reached but was stuck. I'll check a book for formulas then...
    About the rest, I'll try it tomorrow probably and will post any help I feel to ask.
    u + 1 is a linear function, you can integrate with it much the same way you would as with a single variable (just divide by it's derivative when done, which in this case is 1, so it makes no difference). if it helps you to see things clearer, do a secondary substitution, t = u + 1, this yields \int \frac 1{t^2}~dt = \int t^{-2}~dt
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thanks to both I could do them all. I want to notice (for any further reader) that for the 4) \int \frac{dx}{1+e^x}, it's worth \int \frac{du}{u(u-1)} for the u-sub u=e^x+1. Using partial fractions as said bobak we obtain the final result which is worth x-ln(e^x+1)+C.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    It's also worth to see that \frac{1}{1+e^{x}}=\frac{1+e^{x}-e^{x}}{1+e^{x}}=1-\frac{e^{x}}{1+e^{x}}, and by integrating those are easy, no substitution needed.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] substitution indefinite integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 19th 2009, 12:52 PM
  2. Substitution Integrals.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 18th 2009, 08:52 PM
  3. integrals by substitution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 11th 2008, 09:25 PM
  4. U - Substitution for integrals
    Posted in the Calculus Forum
    Replies: 13
    Last Post: June 13th 2008, 10:25 PM
  5. Integrals using U-Substitution
    Posted in the Calculus Forum
    Replies: 18
    Last Post: February 2nd 2008, 01:59 PM

Search Tags


/mathhelpforum @mathhelpforum