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Math Help - Integral

  1. #1
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    Integral

    Evaluate:

    \displaystyle \int^{\frac{\pi}{2}}_0 [\frac{d}{dx} (\sin(\frac{x}{2}) \cos(\frac{x}{3})]dx

    Here is what I did:

    \displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2}) \cos(\frac{x}{3})dx + \int^{\frac{\pi}{2}}_0 \sin(\frac{x}{2})(-\sin(\frac{x}{3}))dx (using the product rule to evaluate d/dx and separating the resultant integrals)

    \displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2})  \cos(\frac{x}{3})dx - \int^{\frac{\pi}{2}}_0  \sin(\frac{x}{2})\sin(\frac{x}{3})dx (just rewriting to pull out the minus sign)

    Assuming that this is right so far, I don't know how to evaluate these two integrals. I'm thinking that maybe I can use FTC to simplify this process somehow?

    Thanks.
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  2. #2
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    hmmmm,

    consider \displaystyle \frac{d}{dx}\left(\sin\left(\frac{x}{2}\right)\rig  ht) =\frac{1}{2}\cos\left(\frac{x}{2}\right)
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  3. #3
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    You're right, but I'm still stuck. I just forgot to apply the chain rule on the derivatives, but all that does is adds a constant to each integral (I think):

    \displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2}) \cos(\frac{x}{3})(\frac{1}{2})dx + \int^{\frac{\pi}{2}}_0 \sin(\frac{x}{2})(-\sin(\frac{x}{3}))(\frac{1}{3})dx

    \displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2}) \cos(\frac{x}{3})(\frac{1}{2})dx - \int^{\frac{\pi}{2}}_0 \sin(\frac{x}{2})\sin(\frac{x}{3})(\frac{1}{3})dx

    I don't see a U substitution here. Am I approaching this correctly?
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  4. #4
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    Isn't the integral of the derivative of a function just that function itself?
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  5. #5
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    Quote Originally Posted by joatmon View Post

    I don't see a U substitution here. Am I approaching this correctly?

    You can continue this way, no u-sub needed, just use,

    \displaystyle \sin a \sin b = \frac{1}{2}(\cos (a+b)-\cos (a-b))

    \displaystyle \cos a \cos b = \frac{1}{2}(\cos (a+b)+\cos (a-b))

    but it should match up with what Hayden has said in post #4
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  6. #6
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    That's what I was wondering, but I am used to seeing it the other way around (with the derivative on the outside), so I didn't think that was legal. Thanks!
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  7. #7
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    Sorry, but I am still confused. If the integral and the derivative "cancel each other out", then do I just evaluate the function at the upper limit like this:

    \displaystyle \sin(\frac{\pi}{4}) \cos(\frac{\pi}{6})
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  8. #8
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    I got it. That's the answer. To finish this off:

    \displaystyle \sin(\frac{\pi}{4}) \cos(\frac{\pi}{6})

    \displaystyle \frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{4}

    Thanks, everyone.
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