Results 1 to 10 of 10

Math Help - Integration questions

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Integration questions

    Hi
    Three questions i am having trouble integrating:
    1) \int{\frac{cosx}{sinx}} dx
    This is what i have done:
    \int{\frac{1}{tanx}}
    so i got:
    sec^2x + c

    answers says its:  ln(sinx)+c

    Someone tell me why is my one incorrect?

    2) \int{\frac{x+1}{x^2+2x}} dx
    Need help how to start this.

    3) \int{\frac{1}{\sqrt{1-9x^2}}}
    This is what i got: sin^{-1}3x+c

    And this what the answers says: \frac{1}{3}sin^{-1}3x+c

    Someone tell me why is my one incorrect?

    P.S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    #1: How did you get \sec^2 x ? Use the substitution: u = \sin x \ \Rightarrow \ du = \cos x \ dx to get a standard integral

    #2: Use the substitution  u = x^2 + 2x \ \Rightarrow \ du = 2(x+1)dx to get another standard integral

    #3: How did you do this question? It looks like you simply used a formula. So if you have a table of integrals, you should see that: \int \frac{1}{\sqrt{1 - (ax)^2} }\ dx \ = \ \frac{1}{a} \arcsin (ax) + C
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2008
    Posts
    509
    Quote Originally Posted by o_O View Post
    #1: How did you get \sec^2 x ? Use the substitution: u = \sin x \ \Rightarrow \ du = \cos x \ dx to get a standard integral
    ok i'll try that.

    Quote Originally Posted by o_O View Post
    #2: Use the substitution  u = x^2 + 2x \ \Rightarrow \ du = 2(x+1)dx to get another standard integral
    i'll try that.

    Quote Originally Posted by o_O View Post
    #3: How did you do this question? It looks like you simply used a formula. So if you have a table of integrals, you should see that: \int \frac{1}{\sqrt{1 - (ax)^2} }\ dx \ = \ \frac{1}{a} \arcsin (ax) + C
    ok i understand now.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008
    Quote Originally Posted by Paymemoney View Post
    3) \int{\frac{1}{\sqrt{1-9x^2}}}
    This is what i got: sin^{-1}3x+c

    And this what the answers says: \frac{1}{3}sin^{-1}3x+c
    If you don't want to use a formula, then you can use trigonometric substitution.

    Notice that

    \int{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int{\frac{1}{\sqrt{9\left(\frac{1}{9} - x^2\right)}}\,dx}

     = \int{\frac{1}{3\sqrt{\frac{1}{9} - x^2}}\,dx}

     = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - x^2}}\,dx}.


    Now make the substitution x = \frac{1}{3}\sin{\theta} so that dx = \frac{1}{3}\cos{\theta}\,d\theta.

    Also notice that \sin{\theta} = 3x and so \theta = \arcsin{3x}.


    Then the integral becomes

     = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - \left(\frac{1}{3}\sin{\theta}\right)^2}}\,\frac{1}  {3}\cos{\theta}\,d\theta}

     = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1}  {9} - \frac{1}{9}\sin^2{\theta}}}\,d\theta}

     = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1}  {9}(1 - \sin^2{\theta})}}\,d\theta}

     = \frac{1}{9}\int{\frac{\cos{\theta}}{\frac{1}{3}\sq  rt{\cos^2{\theta}}}\,d\theta}

     = \frac{1}{3}\int{\frac{\cos{\theta}}{\cos{\theta}}\  ,d\theta}

     = \frac{1}{3}\int{1\,d\theta}

     = \frac{1}{3}\theta + C

     = \frac{1}{3}\arcsin{3x} + C.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2008
    Posts
    509
    Quote Originally Posted by Prove It View Post
    If you don't want to use a formula, then you can use trigonometric substitution.

    Notice that

    \int{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int{\frac{1}{\sqrt{9\left(\frac{1}{9} - x^2\right)}}\,dx}

     = \int{\frac{1}{3\sqrt{\frac{1}{9} - x^2}}\,dx}

     = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - x^2}}\,dx}.


    Now make the substitution x = \frac{1}{3}\sin{\theta} so that dx = \frac{1}{3}\cos{\theta}\,d\theta.

    Also notice that \sin{\theta} = 3x and so \theta = \arcsin{3x}.


    Then the integral becomes

     = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - \left(\frac{1}{3}\sin{\theta}\right)^2}}\,\frac{1}  {3}\cos{\theta}\,d\theta}

     = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1}  {9} - \frac{1}{9}\sin^2{\theta}}}\,d\theta}

     = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1}  {9}(1 - \sin^2{\theta})}}\,d\theta}

     = \frac{1}{9}\int{\frac{\cos{\theta}}{\frac{1}{3}\sq  rt{\cos^2{\theta}}}\,d\theta}

     = \frac{1}{3}\int{\frac{\cos{\theta}}{\cos{\theta}}\  ,d\theta}

     = \frac{1}{3}\int{1\,d\theta}

     = \frac{1}{3}\theta + C

     = \frac{1}{3}\arcsin{3x} + C.
    using the u-substitution looks like i longer way of working out it.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2008
    Posts
    509
    This is what i done in the second question. Can you tell me where i have gone wrong, note this is my first time doing u-substitution.

    u=sinx

    du=cosx

    \int{\frac{cosx}{u}}

    cosx\int{u}

    =cosx \frac{u^2}{2} +c

    =\frac{cos(x)u^2}{2}

    substitute sinx back into u <br /> <br />
=\frac{cosxsinx^2}{2}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2009
    Posts
    226
    You can't factor a variable term like \cos{x} from an integral. Instead of factoring it from the integral, replace it with du because, as you stated, du = \cos{x}. Then, use the integration rule \int \frac{du}{u} = \ln|u| + C.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Ted
    Ted is offline
    Member
    Joined
    Feb 2010
    From
    China
    Posts
    193
    Quote Originally Posted by Paymemoney View Post
    This is what i done in the second question. Can you tell me where i have gone wrong, note this is my first time doing u-substitution.

    u=sinx

    du=cosx

    \int{\frac{cosx}{u}}

    cosx\int{u}

    =cosx \frac{u^2}{2} +c

    =\frac{cos(x)u^2}{2}

    substitute sinx back into u <br /> <br />
=\frac{cosxsinx^2}{2}
    For \int \frac{cos(x)}{sin(x)} dx :

    substitute u=sin(x) to get du=cos(x)dx

    the integral becomes: \int \frac{du}{u}=ln|u|+C=ln|sin(x)|+C

    you should review the substitution method in your book.


    T
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Dec 2008
    Posts
    509
    ok thanks everyone for the help i understand how to do it now.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008
    Quote Originally Posted by Paymemoney View Post
    using the u-substitution looks like i longer way of working out it.
    I'm just showing you how the standard integral was found in the first place - i.e. through trigonometric substitution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Some integration questions
    Posted in the Calculus Forum
    Replies: 11
    Last Post: January 10th 2010, 10:12 AM
  2. Integration questions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 31st 2009, 01:48 PM
  3. Replies: 4
    Last Post: September 16th 2009, 06:01 AM
  4. Two integration questions.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 11th 2008, 10:16 AM
  5. Some integration questions
    Posted in the Calculus Forum
    Replies: 6
    Last Post: August 6th 2008, 11:18 PM

Search Tags


/mathhelpforum @mathhelpforum