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Math Help - trig intergrals

  1. #1
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    [SOLVED] trig intergrals

    So, this is also probably one of those easy problems, I just don't have a good example to work this out by.
    <br />
\int\cos^2x
    where b= \pi/2 a= 0

    answer= \pi/4

    I know I'm suppose to use the half angle identity, i just must be doing it wrong because that's not the answer I'm getting.

    Thanks for all your help ^.^
    Last edited by mortalapeman; August 29th 2008 at 12:01 AM. Reason: SOLVED
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  2. #2
    o_O
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    Can you show your work to see what it is that you're doing wrong?

    \int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx = {\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots
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  3. #3
    Super Member 11rdc11's Avatar
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    cos^2x = \frac{1 + cos2x}{2}

    \frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4}

    Hope this helps.

    Oops was to slow lol, just wondering how do you place upper and lower bounds in latex?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    cos^2x = \frac{1 + cos2x}{2}

    \frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4}

    Hope this helps.

    Oops was to slow lol, just wondering how do you place upper and lower bounds in latex?
    Do:

    Code:
    [tex]\int_a^b f(x)\,dx[/tex]
    To get \int_a^b f(x)\,dx

    note if we have limits like -\ln2 to \ln2, then you need to put these within {} such as:

    Code:
    [tex]\int_{-\ln(2)}^{\ln2} e^x\,dx[/tex]
    will output \int_{-\ln(2)}^{\ln2} e^x\,dx

    --Chris
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  5. #5
    o_O
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    \int_{0}^{4} x \: dx

    is produced by: [tex]\int_{0}^{4} x \: dx[/tex]

    _{0} : produces the lower bound
    ^{4} : produces the upper bound.

    This can applied to other symbols as well:
    \sum_{k = 0}^{4} k = 10 is produced by [tex]\sum_{k = 0}^{4} k = 10[/tex]

    \frac{x}{2}\bigg|_{0}^{4}

    etc etc.
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  6. #6
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    Quote Originally Posted by o_O View Post
    Can you show your work to see what it is that you're doing wrong?

    \int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx = {\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots
    Whoa, I totally was thinking that the anti derivative of 1 was 0. That's where I messed up. Silly mistake. Thanks though, I'll remember to post my work next time.
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  7. #7
    Super Member wingless's Avatar
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    \int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx = {\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots
    There's an easier way to deal with these kind of integrals.

    Rule: \int_a^b f(x)~dx = \int_a^b f(a+b-x)~dx

    I'll solve this step by step for you to understand how it works. It actually takes only a nanosecond to solve this.

    I = \int_0^\frac{\pi}{2}\cos^2x~dx = \int_0^\frac{\pi}{2}\cos^2(\frac{\pi}{2}-x)~dx = \int_0^\frac{\pi}{2}\sin^2x~dx

    Now, I+I = \int_0^\frac{\pi}{2}\cos^2x~dx + \int_0^\frac{\pi}{2}\sin^2x~dx = \int_0^{\frac{\pi}{2}}1~dx = \frac{\pi}{2}

    I = \frac{\pi}{4}
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