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Math Help - 4 integration problems, just need the steps

  1. #1
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    Question 4 integration problems, just need the steps

    Can someone show me the steps to get to the answers?

    x*sin(2*x) dx;
    The answer is:
    1/4*sin(2*x)-1/2*x*cos(2*x)

    1/(x^2*(x^2+4)) dx;
    The answer is -1/8*arctan(1/2*x)-1/(4*x)

    (sin*(x))/(1+cos^2*(x)) dx;
    The answer is:
    sin/cos^2*x-sin/cos^4*ln(1+cos^2*x)

    x/(sqrt(9-x^2)) dx 0to1
    The answer is:
    1/(3+2*2^(1/2))
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  2. #2
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    Quote Originally Posted by LightEight View Post
    Can someone show me the steps to get to the answers?

    x*sin(2*x) dx;
    The answer is:
    1/4*sin(2*x)-1/2*x*cos(2*x)
    Use itegration by parts.

    \int u'v = uv - \int v'u

    where

    v = x \Rightarrow v' = 1

    u' = \sin(2x) \Rightarrow u = -\frac{1}{2}\cos(2x)


    Now use \int u'v = uv - \int v'u to finish it.

    Shouldn't be too tricky.
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  3. #3
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    Quote Originally Posted by LightEight View Post
    Can someone show me the steps to get to the answers?

    x*sin(2*x) dx;
    The answer is:
    1/4*sin(2*x)-1/2*x*cos(2*x)

    1/(x^2*(x^2+4)) dx;
    The answer is -1/8*arctan(1/2*x)-1/(4*x)

    (sin*(x))/(1+cos^2*(x)) dx;
    The answer is:
    sin/cos^2*x-sin/cos^4*ln(1+cos^2*x)

    x/(sqrt(9-x^2)) dx 0to1
    The answer is:
    1/(3+2*2^(1/2))
    You should try to show some effort. You're not suppose to just post problem after problem for others to solve. I don't think there's any rule against posting more then one problem, but you really should show that you've attempted to do this yourself. That way we know what topic you need to work on. However, I'll help get you started with the first integral.

    I(x)=\int xsin(2x)dx

    I'll use a substitution to modify it.

    Let u=2x so that du=2dx. The integral becomes \int\frac{1}{2}usin(u)\frac{du}{2}=\frac{1}{4}\int usin(u)du.

    Now use integration by parts to finish. The formula is:

    \int f(u)g'(u)du=f(u)g(u)-\int g(u)f'(u)du

    Let f(u)=u and g'(u)=sin(u).

    Now see if you can use this information to find  \frac{1}{4}\int usin(u)du.
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  4. #4
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    Thanks, I was able to do them now.
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