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Math Help - Integration

  1. #1
    Member Furyan's Avatar
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    Integration

    Hello

    Can someone help me with this please:

    \int\dfrac{3}{2}\cos^2(4x)

    I tried a number of things, but can't see what function differentiates to give \cos^2(4x)

    A hint at the right approach would be very much appreciated.

    Thank you.
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  2. #2
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    Re: Integration

    Hello, Furyan!

    Can someone help me with this please?

    . . \int\tfrac{3}{2}\cos^2(4x)

    You need this identity: . \cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}

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  3. #3
    Super Member ILikeSerena's Avatar
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    Re: Integration

    Hi Furyan!

    Are you aware of the double angle formula for \cos(2y) in terms of \cos^2y?
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  4. #4
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    Re: Integration

    Quote Originally Posted by Furyan View Post
    Hello

    Can someone help me with this please:

    \int\dfrac{3}{2}\cos^2(4x)

    I tried a number of things, but can't see what function differentiates to give \cos^2(4x)

    A hint at the right approach would be very much appreciated.

    Thank you.
    It would help if you realise that \displaystyle \begin{align*} \cos^2{X} \equiv \frac{1}{2} + \frac{1}{2}\cos{2X} \end{align*}...
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  5. #5
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    Re: Integration

    Quote Originally Posted by Soroban View Post
    Hello, Furyan!


    You need this identity: . \cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}

    You don't really NEED this identity, though it is the simplest simplification.

    Integration by Parts will also work...
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  6. #6
    Newbie MarceloFantini's Avatar
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    Re: Integration

    The following identity may be useful:

    \cos^2 x = \frac{1 + \cos 2x}{2}

    Try using it.
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  7. #7
    Member Furyan's Avatar
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    Re: Integration

    Hello Soroban,

    Quote Originally Posted by Soroban View Post
    Hello, Furyan!


    You need this identity: . \cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}

    Thank you. I was able to find the integral using that identity and substituting \cos^2 4\theta with \dfrac{1}{2}(1 + \cos8\theta)
    Last edited by Furyan; December 30th 2011 at 09:28 AM. Reason: Correction
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  8. #8
    Member Furyan's Avatar
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    Re: Integration

    Hello Prove It

    Quote Originally Posted by Prove It View Post
    It would help if you realise that \displaystyle \begin{align*} \cos^2{X} \equiv \frac{1}{2} + \frac{1}{2}\cos{2X} \end{align*}...
    Thank you. I am having trouble with that realisation but I'm working on it.

    I will also try using integration by parts.
    Last edited by Furyan; December 30th 2011 at 08:29 AM. Reason: Addition
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  9. #9
    Newbie MarceloFantini's Avatar
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    Re: Integration

    Do you mean \cos^2 4x = \frac{1 + \cos 8x}{2} instead of \cos^2 4x = \frac{1+8 \cos x}{2}?
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  10. #10
    Member Furyan's Avatar
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    Re: Integration

    MarceloFantini

    Quote Originally Posted by MarceloFantini View Post
    Do you mean \cos^2 4x = \frac{1 + \cos 8x}{2} instead of \cos^2 4x = \frac{1+8 \cos x}{2}?
    Yes indeed, thank you for pointing that out. I have corrected my post accordingly. I did use the former when I found the integral.
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  11. #11
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    Re: Integration

    Quote Originally Posted by Furyan View Post
    Hello Prove It



    Thank you. I am having trouble with that realisation but I'm working on it.

    I will also try using integration by parts.
    Here's how that identity is derived.

    First a proof of the angle sum formula \displaystyle \begin{align*} \cos{\left(\alpha + \beta\right)} \equiv \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \end{align*}

    Ans 1

    Now, using the angle sum formula with \displaystyle \begin{align*} \alpha = \beta = X \end{align*} we have

    \displaystyle \begin{align*} \cos{\left(X + X\right)} &\equiv \cos{(X)}\cos{(X)} - \sin{(X)}\sin{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \sin^2{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \left[1 - \cos^2{(X)}\right] \textrm{ by the Pythagorean Identity} \\ \cos{(2X)} &\equiv 2\cos^2{(X)} - 1 \\ 2\cos^2{(X)} &\equiv 1 + \cos{(2X)} \\ \cos^2{(X)} &\equiv \frac{1}{2} + \frac{1}{2}\cos{(2X)} \end{align*}


    And if you're trying to use integration by parts...

    \displaystyle \begin{align*} I &= \int{\frac{3}{2}\cos^2{(4x)}\,dx} \\ I &= \frac{3}{2}\int{\cos{(4x)}\cos{(4x)}\,dx} \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} - \int{-4\sin{(4x)}\cdot \frac{1}{4}\sin{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{\sin^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1 - \cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1\,dx} - \int{\cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{8} \sin{(4x)} \cos{(4x)} + \frac{3}{2} \int{1\,dx} - \frac{3}{2} \int{ \cos^2{ (4x) }\,dx} \\ I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x - I \\ 2I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x \\ I &= \frac{3}{16}\sin{(4x)}\cos{(4x)} + \frac{3}{4}x + C  \end{align*}

    And with a little manipulation I'm sure you can get this to be the same form as your answer you gained using the double angle identity...
    Last edited by Prove It; December 30th 2011 at 04:55 PM.
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  12. #12
    Member Furyan's Avatar
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    Re: Integration

    Dear Prove It

    Quote Originally Posted by Prove It View Post
    Here's how that identity is derived.

    First a proof of the angle sum formula \displaystyle \begin{align*} \cos{\left(\alpha + \beta\right)} \equiv \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \end{align*}

    Ans 1

    Now, using the angle sum formula with \displaystyle \begin{align*} \alpha = \beta = X \end{align*} we have

    \displaystyle \begin{align*} \cos{\left(X + X\right)} &\equiv \cos{(X)}\cos{(X)} - \sin{(X)}\sin{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \sin^2{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \left[1 - \cos^2{(X)}\right] \textrm{ by the Pythagorean Identity} \\ \cos{(2X)} &\equiv 2\cos^2{(X)} - 1 \\ 2\cos^2{(X)} &\equiv 1 + \cos{(2X)} \\ \cos^2{(X)} &\equiv \frac{1}{2} + \frac{1}{2}\cos{(2X)} \end{align*}


    And if you're trying to use integration by parts...

    \displaystyle \begin{align*} I &= \int{\frac{3}{2}\cos^2{(4x)}\,dx} \\ I &= \frac{3}{2}\int{\cos{(4x)}\cos{(4x)}\,dx} \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} - \int{-4\sin{(4x)}\cdot \frac{1}{4}\sin{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{\sin^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1 - \cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1\,dx} - \int{\cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{8} \sin{(4x)} \cos{(4x)} + \frac{3}{2} \int{1\,dx} - \frac{3}{2} \int{ \cos^2{ (4x) }\,dx} \\ I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x - I \\ 2I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x \\ I &= \frac{3}{16}\sin{(4x)}\cos{(4x)} + \frac{3}{4}x + C  \end{align*}

    And with a little manipulation I'm sure you can get this to be the same form as your answer you gained using the double angle identity...
    I have been looking at the double angle identity for \cos2\theta and, for some reason, was having trouble realising what you had pointed out in your original post. Now it's crystal clear. Some how I keep missing the elementary principles.

    I will try and find the integral again using the substitution you have suggested.

    Thank you also for the help with using integration by parts. I'd never have been able to do that, but I am very interested in it and and will work through it until I understand it.

    Thank you very much for the time you have taken to help me. It's extremely generous of you and very grateful.
    Last edited by Furyan; December 30th 2011 at 05:28 PM. Reason: Correction
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