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Thread: Integration Help Calculus

  1. #1
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    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!
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  2. #2
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    Sorry, do you mean $\displaystyle \int_0^{\pi/2}{(x+\cos{x})\sin{x}}dx$?
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  3. #3
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    Yes, sorry about the confusion. I have no clue how to write the problems in that form.
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  4. #4
    Super Member Aryth's Avatar
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    So, you want to find:

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

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

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

    Now, you can split the integral:

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

    Let's go ahead and do the second integral:

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

    This can be solved with simple substitution:

    $\displaystyle u = sin(x)$

    $\displaystyle du = cos(x)dx$

    Now we redo the limits in terms of u:

    $\displaystyle u(0) = sin(0) = 0$

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

    Our new integral is:

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

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

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

    We have to use parts:

    $\displaystyle u = x$

    $\displaystyle du = dx$

    $\displaystyle dv = sin(x)dx$

    $\displaystyle v = -cos(x)$

    Now, we have a complete integral:

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

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

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

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

    So, the total is:

    $\displaystyle 1 + \frac{1}{2} = \frac{3}{2}$
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    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!
    $\displaystyle \int_{0}^{\frac{\pi}{2}}{\left(x+cosx\right)}sinx\ ,dx$.

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

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

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

    We now have the following integral:

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

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

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

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

    Now we have:

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

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

    Now, we have the following:

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

    Evaluating, we get:

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

    Hope this makes sense! If you want me to clarify, let me know!
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