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Thread: [SOLVED] Trigonometric Integral

  1. #1
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    [SOLVED] Trigonometric Integral

    How is this integrated?

    $\displaystyle \int{\cos^2{\Theta}}d\Theta$
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  2. #2
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    Quote Originally Posted by Mike9182 View Post
    How is this integrated?

    $\displaystyle \int{\cos^2{\Theta}}d\Theta$
    Racall the identity: $\displaystyle cos^2(\theta)=\frac{1}{2}(cos(2\theta)+1)$.
    Hence:
    $\displaystyle \int cos^2(\theta) d\theta = \frac{1}{2} \int cos(2\theta) d\theta + \frac{1}{2} \int d\theta $.

    It should be easy now.
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  3. #3
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    How do you integrate $\displaystyle \frac{1}{2} \int(\cos{2\Theta})d\Theta$?
    Last edited by Mike9182; Jan 26th 2010 at 03:18 PM. Reason: Latex error
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    Quote Originally Posted by Mike9182 View Post
    How do you integrate $\displaystyle \frac{1}{2} \int(\cos{2\Theta})d\Theta$?

    $\displaystyle
    \int(\cos{k\Theta})d\Theta, k\in \mathbb{R} = \frac{1}{k} \sin{k\Theta}+C
    $
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  5. #5
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    Quote Originally Posted by Mike9182 View Post
    How do you integrate $\displaystyle \frac{1}{2} \int(\cos{2\Theta})d\Theta$?
    there is two ways:
    1- Substitute $\displaystyle u=2\theta$.
    2- Apply this formula:
    $\displaystyle b\int cos(a\theta) d\theta = \frac{b}{a} sin(a\theta) + C$ for $\displaystyle a \neq 0 $
    in your question: $\displaystyle a=2$ and $\displaystyle b=\frac{1}{2}$.

    Quote Originally Posted by pickslides View Post
    $\displaystyle
    \int(\cos{k\Theta})d\Theta, k\in \mathbb{R} = \frac{1}{k} \sin{k\Theta}+C
    $
    $\displaystyle k \neq 0$
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  6. #6
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    Thanks, I understand now.
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  7. #7
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    Quote Originally Posted by Mike9182 View Post
    How is this integrated?

    $\displaystyle \int{\cos^2{\Theta}}d\Theta$
    Another way...

    $\displaystyle \int{Cos\theta\ Cos\theta}d\theta=\int{u}dv=uv-\int{v}du$

    $\displaystyle u=Cos\theta,\ \frac{du}{d\theta}=-Sin\theta,\ du=-Sin\theta\ d\theta$

    $\displaystyle dv=Cos\theta\ d\theta,\ v=\int{Cos\theta}d\theta=Sin\theta$

    $\displaystyle uv-\int{v}du=Cos\theta Sin\theta+\int{Sin^2\theta}d\theta$

    $\displaystyle \Rightarrow\ I=Cos\theta Sin\theta+\int{(1-Cos^2\theta)}d\theta=Cos\theta Sin\theta+\int{(1)}d\theta-I$

    $\displaystyle 2I=Cos\theta Sin\theta+\theta\ \Rightarrow\ I=\frac{1}{2}Cos\theta Sin\theta+\frac{\theta}{2}+C=\frac{1}{4}Sin2\theta +\frac{\theta}{2}+C$
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