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Math Help - [SOLVED] integral

  1. #1
    fid
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    [SOLVED] integral

    HI ,

    i need the integral of:

    e^ax * cos(bx) *dx
    x^k * sin(ax) *dx
    x^k * e^ax *dx

    thanks in advance
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fid View Post
    HI ,

    i need the integral of:

    e^ax * cos(bx) *dx
    Assume a,b \neq 0, the solution is trivial in any of the cases a = 0 or b = 0 or both a = b = 0.

    So, if a,b \neq 0 we can proceed by Integration by parts.

    \int e^{ax} \cos (bx) dx = \frac {1}{a}e^{ax} \cos (bx) + \frac {b}{a} \int e^{ax} \sin (bx)dx .......do this last integral by parts again

    \Rightarrow \int e^{ax} \cos (bx)dx = \frac {1}{a}e^{ax} \cos (bx) + \frac {b}{a} \left[ \frac {1}{a}e^{ax} \sin(bx) - \frac {b}{a}\int e^{ax} \cos(bx)dx \right]

    \Rightarrow \int e^{ax} \cos(bx)dx = \frac {1}{a}e^{ax} \cos(bx) + \frac {b}{a^2}e^{ax} \sin(bx) - \left( \frac {b}{a} \right)^2 \int e^{ax} \cos (bx)dx

    \Rightarrow \left[1 + \left( \frac {b}{a} \right)^2 \right] \int e^{ax} \cos (bx)dx = \frac {1}{a}e^{ax} \cos(bx) + \frac {b}{a^2} e^{ax} \sin(bx)

    \Rightarrow \int e^{ax} \cos(bx)dx = \frac { \frac {1}{a}e^{ax} \cos (bx) + \frac {b}{a^2}e^{ax} \sin(bx)}{ \left[1 + \left( \frac {b}{a} \right)^2 \right]}

    Someone please check this. I still get lost when typing out solutions with LaTex

    I see others replying to the thread, so i'll hold off on answering the others. i'd only answer one though, since the last two are kind of similar
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Looks like correct.

    I got

    \int{e^{ax}\cdot\cos{bx}~dx}=\frac{ae^{ax}\cos{bx}  +be^{ax}\sin{bx}}{a^2+b^2}+k
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Looks like correct.

    I got

    \int{e^{ax}\cdot\cos{bx}~dx}=\frac{ae^{ax}\cos{bx}  +be^{ax}\sin{bx}}{a^2+b^2}+k
    yeah, i think that's what mine simplifies to, i could bother simplifying however. And i forgot the arbitrary constant...I always do that!

    did you use a different method?
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  5. #5
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    Hello, fid!

    These are very messy problems . . . Here's the first one.


    I \;= \;\int e^{ax}\cos(bx)\,dx

    Integrate by parts:
    . . \begin{array}{ccccccc}u & = & \cos(bx) & \quad & dv & = & e^{ax}dx \\<br />
du & = & -b\sin(bx)\,dx & \quad & v & = & \frac{1}{a}e^{ax} \end{array}

    We have: . I \;=\;\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\int e^{ax}\sin(bx)\,dx

    . . Integrate by parts:
    . . . . \begin{array}{ccccccc}u & = & \sin(bx) & \quad & dv &= & e^{ax}dx \\<br />
du & = & b\cos(bx)\,dx & \quad & v & = & \frac{1}{a}e^{ax} \end{array}

    We have: . I \;=\;\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left[\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right]

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

    We have: . . I \;=\;\frac{1}{a}e^{ax}\cos(bx) +  \frac{b}{a^2}e^{ax}\sin(bx) - \frac{b^2}{a^2}I

    Then: . I + \frac{b^2}{a^2}I \;=\;\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) + C

    . . . \left(\frac{a^2+b^2}{a^2}\right)I \;=\;\frac{e^{ax}}{a^2}\bigg[a\cos(bx) + b\sin(bx)\bigg] + C

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


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


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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by fid View Post
    HI ,

    i need the integral of:

    e^ax * cos(bx) *dx
    x^k * sin(ax) *dx
    x^k * e^ax *dx

    thanks in advance
    Look up integration by parts. The first one is the trickiest so I'll do that one.

    \int e^{ax}cos(bx) \, dx

    We know that \int u dv = uv - \int v du (from the product rule) so choose:
    u = cos(bx) ==> du = -b \cdot sin(bx) dx
    dv = e^{ax} dx ==> v = \frac{1}{a}e^{ax}

    So:
    \int e^{ax}cos(bx) \, dx = cos(bx) \cdot \frac{1}{a}e^{ax} - \int \frac{1}{a}e^{ax} \cdot -b \, sin(bx) \, dx = \frac{1}{a}e^{ax}cos(bx) + \frac{b}{a} \int e^{ax}sin(bx) \, dx

    Now do \int e^{ax}sin(bx) \, dx.

    Choose
    u = sin(bx) ==> du = b \cdot cos(bx) dx
    dv = e^{ax} dx ==> v = \frac{1}{a}e^{ax}

    Then
    \int e^{ax}sin(bx) \, dx = sin(bx) \cdot \frac{1}{a}e^{ax} - \int \frac{1}{a}e^{ax} \cdot b \cdot cos(bx) dx = <br />
\frac{1}{a}e^{ax}sin(bx) - \frac{b}{a} \int <br />
e^{ax}cos(bx) dx

    Now substitute this value for the sin(bx) integral into your cos(bx) expression:
    \int e^{ax}cos(bx) \, dx = \frac{1}{a}e^{ax}cos(bx) + \frac{b}{a} \int e^{ax}sin(bx) \, dx = \frac{1}{a}e^{ax}cos(bx) + \frac{b}{a} \left ( <br />
\frac{1}{a}e^{ax}sin(bx) - \frac{b}{a} \int <br />
e^{ax}cos(bx) dx \right )

    \int e^{ax}cos(bx) \, dx = \frac{1}{a}e^{ax}cos(bx) + \frac{b}{a^2}e^{ax}sin(bx) - \frac{b^2}{a^2} \int <br />
e^{ax}cos(bx) dx

    This would apparently give us nothing because the solution of the integral depends on the value of the integral! But treat \int e^{ax}cos(bx) \, dx as a variable for the moment. If we add \frac{b}{a^2} \int e^{ax}cos(bx) dx to both sides...

    \int e^{ax}cos(bx) \, dx + \frac{b^2}{a^2} \int <br />
 e^{ax}cos(bx) dx = \frac{1}{a}e^{ax}cos(bx) + \frac{b}{a^2}e^{ax}sin(bx)

    Now factor the common \int e^{ax}cos(bx) \, dx:
    \left ( \int e^{ax}cos(bx) \, dx \right) \left ( 1 + \frac{b^2}{a^2} \right ) = \frac{1}{a}e^{ax}cos(bx) + \frac{b}{a^2}e^{ax}sin(bx)

    \int e^{ax}cos(bx) \, dx = \frac{ \frac{1}{a}e^{ax}cos(bx) + \frac{b}{a^2}e^{ax}sin(bx)}{1 + \frac{b^2}{a^2}}

    \int e^{ax}cos(bx) \, dx = \frac{ ae^{ax}cos(bx) + be^{ax}sin(bx)}{a^2 + b^2}

    -Dan

    (Oops! As Soroban, Krizalid, and Jhevon all said, you need to add an arbitrary constant to this. Wow was I slow on this one! )
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by fid View Post
    x
    x^k * sin(ax) *dx
    x^k * e^ax *dx
    I'm not going to do these last two, but consider using
    u = x^k
    and
    dv = sin(ax)
    for the second one. The third one is done in a similar fashion.

    -Dan
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  8. #8
    Math Engineering Student
    Krizalid's Avatar
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    By the way!!!

    Soroban, did you add me to your msn list?

    I already added you. My msn is asymmetrical_line@hotmail.com

    I hope to see you there!!
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    I'm not going to do these last two, but consider using
    u = x^k
    and
    dv = sin(ax)
    for the second one. The third one is done in a similar fashion.

    -Dan
    the second one is a bit more complicated i believe. we have to deal with factorials and also, i think, we have to do two cases: if k is odd or even. don't think we need cases for the third, but factorials may come into play again

    it seems everyone else simplified their answer. turns out i was the lazy one here...hey, i had to leave something for fid to do on this question, right?
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  10. #10
    Grand Panjandrum
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    For the first a simple way of doing the integral is to observe that:

    \int{e^{ax}\cdot\cos{bx}~dx}=Re\left[\int e^{(a+\bold{i}b)x} ~dx\right]

    RonL
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