Results 1 to 5 of 5

Math Help - Integration Help Calculus

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    8

    Question Integration Help Calculus

    I am new to this forum..so hello everyone! I am a student at UCD and sadly they make me take Calculus, even though I am a Political Science major...So please help me, I do not know if I did this right....

    Find the Integral

    (x + cos x) sin xdx the intervals are 0 and pi/2

    My answer: [(sin x )^2]/2 + C...which is then .5-0 and then .5

    Is this right?

    Thank you so much!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    May 2008
    Posts
    16
    Sorry, do you mean \int_0^{\pi/2}{(x+\cos{x})\sin{x}}dx?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    8
    Yes, sorry about the confusion. I have no clue how to write the problems in that form.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    652
    Thanks
    2
    Awards
    1
    So, you want to find:

    \int_0^{\frac{\pi}{2}} [sin(x)(x + cos(x))]dx

    You can go ahead and distribute the sin(x):

    \int_0^{\frac{\pi}{2}} [xsin(x) + sin(x)cos(x)]dx

    Now, you can split the integral:

    \int_0^{\frac{\pi}{2}} [xsin(x)]dx + \int_0^{\frac{\pi}{2}} [sin(x)cos(x)]dx

    Let's go ahead and do the second integral:

    \int_0^{\frac{\pi}{2}} [sin(x)cos(x)]dx

    This can be solved with simple substitution:

    u = sin(x)

    du = cos(x)dx

    Now we redo the limits in terms of u:

    u(0) = sin(0) = 0

    u\left(\frac{\pi}{2}\right) = sin\left(\frac{\pi}{2}\right) = 1

    Our new integral is:

    \int_0^1 [u]du = \frac{u^2}{2}]^1_0 = \frac{1}{2}

    Now, we have to go back to the first integral:

    \int_0^{\frac{\pi}{2}} [xsin(x)]dx

    We have to use parts:

    u = x

    du = dx

    dv = sin(x)dx

    v = -cos(x)

    Now, we have a complete integral:

    [-xcos(x)]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} [-cos(x)]dx

    Since cos\left(\frac{\pi}{2}\right) = 0 the first term is gone, which leaves the second term, which is:

    \int_0^{\frac{\pi}{2}} [cos(x)]dx <--- Cancelled the negatives

     = [sin(x)]_0^{\frac{\pi}{2}} = sin\left(\frac{\pi}{2}\right) = 1

    So, the total is:

    1 + \frac{1}{2} = \frac{3}{2}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by bloomsgal8 View Post
    I am new to this forum..so hello everyone! I am a student at UCD and sadly they make me take Calculus, even though I am a Political Science major...So please help me, I do not know if I did this right....

    Find the Integral

    (x + cos x) sin xdx the intervals are 0 and pi/2

    My answer: [(sin x )^2]/2 + C...which is then .5-0 and then .5

    Is this right?

    Thank you so much!
    \int_{0}^{\frac{\pi}{2}}{\left(x+cosx\right)}sinx\  ,dx.

    You can't make a direct substitution quite yet. Distribute the sine term throughout the parenthesis to get this:

    \int_{0}^{\frac{\pi}{2}}{\left(xsinx+sinxcosx\righ  t)}\,dx.

    Recall that 2sinxcosx=sin2x. \therefore sinxcosx=\frac{1}{2}sin2x.

    We now have the following integral:

    \int_{0}^{\frac{\pi}{2}}{\left(xsinx\right)}\,dx+\  frac{1}{2}\int_{0}^{\frac{\pi}{2}}sin2x\,dx.

    You will need to apply integration by parts to the first integral, and directly integrate the second.

    Integration by parts : \int u\,dv=uv-\int v\,du.

    let u=x and dv=sinx\,dx.
    \therefore du=dx and v=-cosx.

    Now we have:

    \int xsinx\,dx=-xcosx+\int cosx\,dx
    \therefore \int xsinx\,dx=-xcosx+sinx+C.

    Since we're dealing with definite integrals, we don't need the C.

    Now, we have the following:

    \int_{0}^{\frac{\pi}{2}}{\left(xsinx\right)}\,dx+\  frac{1}{2}\int_{0}^{\frac{\pi}{2}}sin2x\,dx={\left  (-xcosx+sinx-\frac{1}{4}cos2x\right)}_{0}^{\frac{\pi}{2}}.

    Evaluating, we get:

    {\left(-{\left(\frac{\pi}{2}\right)}cos{\left(\frac{\pi}{2  }\right)}+sin{\left(\frac{\pi}{2}\right)}-\frac{1}{4}cos2{\left(\frac{\pi}{2}\right)}\right)  }-{\left(-{\left(0\right)}cos{\left(0\right)}+sin{\left(0\ri  ght)}-\frac{1}{4}cos2{\left(0\right)}\right)}
    ={\left(1+\frac{1}{4}\right)}-{\left(-\frac{1}{4}\right)}=\frac{3}{2}.

    Hope this makes sense! If you want me to clarify, let me know!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calculus - Integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 10th 2010, 04:23 PM
  2. Calculus - Integration
    Posted in the Calculus Forum
    Replies: 18
    Last Post: January 10th 2010, 05:24 AM
  3. [Calculus AB] Integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 24th 2009, 03:25 AM
  4. Calculus 2 Integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 17th 2008, 01:28 PM
  5. calculus integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 8th 2008, 06:32 PM

Search Tags


/mathhelpforum @mathhelpforum