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Math Help - Help Calculas

  1. #1
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    Solved : Help Calculas

    Thanks guys for the help provided.
    Last edited by tester85; October 24th 2008 at 07:36 PM.
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  2. #2
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    Hello, tester85!

    I'll explain the first one for you.
    It's quite challenging and uses a technique you may not have seen.


    I \;=\;\int e^{ax}\sin bx\,dx
    Integrate by parts . . .

    . . \begin{array}{ccccccc}u &=& \sin bx & & dv &=& e^{ax}dx \\ du &=& b\cos bx\,dx & & v &=&\frac{1}{a}e^{ax} \end{array}

    We have: . I \;=\;\tfrac{1}{a}\,e^{ax}\sin bx - \tfrac{b}{a}\int\! e^{ax}\cos bx\,dx


    Integrate by parts (again):

    . . \begin{array}{ccccccc}u &=&\cos bx & & dv &=& e^{ax}dx \\ du &=& \text{-}b\sin bx\,dx & & v &=& \frac{1}{a}e^{ax} \end{array}

    We have: . I \;=\;\tfrac{1}{a}\,e^{ax}\sin bx - \tfrac{b}{a}\bigg[\tfrac{1}{a}e^{ax}\cos bx + \tfrac{b}{a}\int\! e^{ax}\sin bx\,dx\bigg] + C

    . . I \;=\;\tfrac{1}{a},e^{ax}\sin bx - \tfrac{b}{a^2}\,e^{ax}\cos bx - \tfrac{b^2}{a^2}\underbrace{\int\! e^{ax}\sin bx\,dx}_{\text{This is }I} + C

    . . I \;=\;\tfrac{1}{a}\,e^{ax}\sin bx - \tfrac{b}{a^2}\,e^{ax}\cos bx - \tfrac{b^2}{a^2}\,I + C


    Multiply by a^2\!:\;\;a^2I \;=\;ae^{ax}\sin bx - be^{ax}\cos bx - b^2I + C

    . . . . . . . . a^2I + b^2I \;=\;ae^{ax}\sin bx - be^{ax}\cos bx + C


    Factor: . (a^2+b^2)I \;=\;e^{ax}(a\sin bx - b\cos bx) + C

    . . . . . . . . . . . I \;=\;\frac{e^{ax}}{a^2+b^2}(a\sin bx - b\cos bx) + C


    Therefore: . \int e^{ax}\sin bx\,dx \;=\;\frac{e^{ax}}{a^2+b^2}(a\sin bx - b\cos bx) + C

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  3. #3
    Senior Member Peritus's Avatar
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    alternatively you can use the well known identity:

    <br />
\sin bx = \frac{{e^{jbx}  - e^{ - jbx} }}<br />
{{2j}}

    so the integral becomes a simple combination of exponents:


    <br />
\int {e^{ax} \sin bxdx = \frac{1}<br />
{{2j}}\int {e^{x(a + jb)}  - e^{x(a - jb)} dx} } <br />

    \frac{1}<br />
{{2j}}\int {e^{x(a + jb)}  - e^{x(a - jb)} dx}  = \frac{1}<br />
{{2j}}\left[ {\frac{1}<br />
{{a + jb}}e^{x(a + jb)}  - \frac{1}<br />
{{a - jb}}e^{x(a - jb)} } \right] = <br />

    <br />
 = \frac{1}<br />
{{2j}}e^{ax} \left[ {\frac{{ae^{jbx}  - jbe^{jbx}  - ae^{ - jbx}  - jbe^{ - jbx} }}<br />
{{a^2  + b^2 }}} \right] = \frac{{e^{ax} }} {{a^2  + b^2 }}\left( {a\sin bx - b\cos bx} \right)<br />
    Last edited by Peritus; October 24th 2008 at 09:58 AM.
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  4. #4
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    Thanks Soroban and peritus. Can someone guide me with the rest, i have done it but unable to get the final solution.
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  5. #5
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    \int{\frac{x^2}{(x-1)(x+1)^2}\,dx}

    Use partial fractions.

     \frac{x^2}{(x-1)(x+1)^2} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2}

     = \frac{A(x+1)^2 + B(x-1)(x+1) + C(x-1)}{(x-1)(x+1)^2}

    Since the denominators are the same, the numerators are equal...

    So x^2 = A(x+1)^2 + B(x-1)(x+1) + C(x-1)

    Expand and you get

    x^2 + 0x + 0 = (A + B)x^2 + (2A + C)x + A - B - C

    Equating like powers of x gives

    A + B = 1, 2A + C = 0, A - B - C = 0.

    Solving the equations simultaneously gives

    A = \frac{1}{4}, B = \frac{3}{4}, C = -\frac{1}{2}.


    So \frac{x^2}{(x-1)(x+1)^2} = \frac{1}{4} \left(\frac{1}{x-1}\right) + \frac{3}{4} \left(\frac{1}{x+1}\right) - \frac{1}{2} \left(\frac{1}{(x+1)^2}\right)


    \int{\frac{x^2}{(x-1)(x+1)^2}\,dx} = \frac{1}{4} \int{\frac{1}{x-1}\,dx} + \frac{3}{4} \int{\frac{1}{x+1}\,dx} - \frac{1}{2} \int{\frac{1}{(x+1)^2}\,dx}

     = \frac{1}{4} \ln |x-1| + \frac{3}{4} \ln |x + 1| + \frac{1}{2(x+1)} + c
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  6. #6
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    Hi prove it.

    I got the same answer as what you got, but the answer given is

    1/4 ln|(x-1)(x+1)^3|+1/2(x+1)+C
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  7. #7
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    Quote Originally Posted by tester85 View Post
    Hi prove it.

    I got the same answer as what you got, but the answer given is

    1/4 ln|(x-1)(x+1)^3|+1/2(x+1)+C
    They've used the log laws...

    \log m + \log n = \log mn and p\log m = \log{m^p}
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  8. #8
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    Ok now i get it thanks.

    Therefore, 3/4ln(x+1)=1/4ln(x+1)^3
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  9. #9
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    For the last one, you need to set up the partial fractions as

    \frac{A}{x+1} + \frac{Bx+C}{x^2 + 1}.

    You should find A = \frac{1}{2}, B = -\frac{1}{2}, C = \frac{1}{2}.


    Knowledge that \frac{-x + 1}{x^2 + 1} = \frac{-x}{x^2 + 1} + \frac{1}{x^2 + 1} should also help you out.
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  10. #10
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    I am able to do the partial fractions part but the final answer is wierd, it is given as 1/4ln2 + pi/8 . Where did the pi come from ?
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  11. #11
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    Quote Originally Posted by tester85 View Post
    I am able to do the partial fractions part but the final answer is wierd, it is given as 1/4ln2 + pi/8 . Where did the pi come from ?
    You'll get an inverse trig function in your answer.
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  12. #12
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    Ok noted with thanks prove it. Will try again later. Does anyone has any tips for the second integration question ? The answer given is 4 and i am not able to get it.
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  13. #13
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    That depends.

    Is it...

    \sqrt{1 - \cos{\frac{x}{2}}} or \sqrt{1 - \frac{\cos{x}}{2}} or \sqrt{\frac{1 - \cos{x}}{2}}?
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  14. #14
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by tester85 View Post
    integrate from 2pi to 0 sqrt(1-cosx/2)
    I'd like to help, but I'm not sure if my interpretation is correct. Do you mean

    \int_0^{2\pi}\sqrt{\frac{1-\cos x}{2}}\,dx or

    \int_0^{2\pi}\sqrt{1-\frac{\cos x}{2}}\,dx or

    \int_0^{2\pi}\sqrt{1-\cos \left(\frac{x}{2}\right)}\,dx??

    --Chris
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  15. #15
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    Quote Originally Posted by Chris L T521 View Post
    I'd like to help, but I'm not sure if my interpretation is correct. Do you mean

    \int_0^{2\pi}\sqrt{\frac{1-\cos x}{2}}\,dx or

    \int_0^{2\pi}\sqrt{1-\frac{\cos x}{2}}\,dx or

    \int_0^{2\pi}\sqrt{1-\cos \left(\frac{x}{2}\right)}\,dx??

    --Chris
    It is this bro. \int_0^{2\pi}\sqrt{\frac{1-\cos x}{2}}\,dx
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