Results 1 to 4 of 4

Math Help - Integration by substitution and part

  1. #1
    Newbie
    Joined
    Feb 2007
    Posts
    22

    Integration by substitution and part

    okay i got two question which i have real problems with. Any help would be great cheers guys

    a)intergrate this using the substitution u = x^2 + 1

    I = ∫2x(x^2 + 1)^3 dx

    b) intergrate this using parts

    J = ∫3xsinx dx

    any help would be brilliant cheers guys
    Last edited by mxmadman_44; August 17th 2007 at 09:38 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by mxmadman_44 View Post
    okay i got two question which i have real problems with. Any help would be great cheers guys

    a)intergrate this using the substitution u = x^2 + 1

    I = ∫2x(x^2 + 1)^3 dx
    Well... u = x^2 + 1 \implies du = 2xdx

    So
    I = \int (x^2 + 1)^3 \cdot 2x dx = \int u^3 du

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by mxmadman_44 View Post
    b) intergrate this using parts

    J = ∫3xsinx dx

    any help would be brilliant cheers guys
    I = 3\int x sin(x) dx

    Let
    u = x \implies du = dx
    dv = sin(x) dx \implies v = -cos(x)

    So
    I = 3\int x sin(x) dx = 3 \left ( -xcos(x) + \int cos(x) dx \right )

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by mxmadman_44 View Post
    b) intergrate this using parts

    J = ∫3xsinx dx
    Here's another solution (I'm not gonna take the constant):

     <br />
\begin{aligned}<br />
\int {x\sin x\,dx} &= \int {\left( {\cos x - \cos x + x\sin x} \right)\,dx}\\<br />
&= \int {\left[ {\cos x - \left( {\cos x - x\sin x} \right)} \right]\,dx}\\<br />
&= \int {\left( {\sin x - x\cos x} \right)'\,dx}\\<br />
&= \sin x - x\cos x + k<br />
\end{aligned}<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. twice integration by part?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 24th 2011, 09:03 PM
  2. Integration with a 2 part du
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 17th 2009, 04:34 AM
  3. integration by part
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 10th 2009, 01:21 PM
  4. integration by part
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 27th 2008, 03:01 PM
  5. integration part 2
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 26th 2008, 12:20 PM

Search Tags


/mathhelpforum @mathhelpforum