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Math Help - Definite Integral problem

  1. #1
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    Definite Integral problem

    <br />
\displaystyle\int_{6}^{13} \sin^2( 4 x ) \cos^2 (4 x) \,dx<br />

    Can someone please point out where I went wrong in my work?
    Thank you

    First, I rewrote this problem as:
    <br />
\displaystyle\int\frac{1}{2}[1-\cos(8x)]\frac{1}{2}[1+\cos(8x)]\,dx<br />
    <br />
\displaystyle\frac{1}{4}\int(1-\cos(8x))(1+\cos(8x))\,dx = \frac{1}{4}\int1-\cos^2(8x)\,dx<br />
    <br />
\displaystyle\frac{1}{4}\int1-\frac{1}{2}[1+\cos(16x)]\,dx=\frac{1}{4}\int\frac{1-cos(16x)}{2}\,dx<br />
    <br />
\displaystyle\frac{1}{8}\int1-\cos(16x)\,dx<br />
    evaluates out to be:
    <br />
\displaystyle\frac{x}{8}-\frac{\sin(16x)}{128}<br />

    Now, when I plug in the upper & lower limits, I get:
    .886437448892

    as a final answer. This does not appear to be correct. What am I doing wrong?

    Thanks!
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  2. #2
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    Quote Originally Posted by Vamz View Post
    <br />
\displaystyle\int_{6}^{13} \sin^2( 4 x ) \cos^2 (4 x) \,dx<br />

    Can someone please point out where I went wrong in my work?
    Thank you

    First, I rewrote this problem as:
    <br />
\displaystyle\int\frac{1}{2}[1-\cos(8x)]\frac{1}{2}[1+\cos(8x)]\,dx<br />
    <br />
\displaystyle\frac{1}{4}\int(1-\cos(8x))(1+\cos(8x))\,dx = \frac{1}{4}\int1-\cos^2(8x)\,dx<br />
    <br />
\displaystyle\frac{1}{4}\int1-\frac{1}{2}[1+\cos(16x)]\,dx=\frac{1}{4}\int\frac{1-cos(16x)}{2}\,dx<br />
    <br />
\displaystyle\frac{1}{8}\int1-\cos(16x)\,dx<br />
    evaluates out to be:
    <br />
\displaystyle\frac{x}{8}-\frac{\sin(16x)}{128}<br />

    Now, when I plug in the upper & lower limits, I get:
    .886437448892

    as a final answer. This does not appear to be correct. What am I doing wrong?

    Thanks!
    your integral is fine. I evaluate it as 0.8779... change your mode to radians.
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  3. #3
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    Let's see...

    \displaystyle \int{\sin^2{(4x)}\cos^2{(4x)}\,dx} = \int{[\sin{(4x)}\cos{(4x)}]^2\,dx}

    \displaystyle = \int{\left[\frac{1}{2}\sin{(8x)}\right]^2\,dx}

    \displaystyle = \int{\frac{1}{4}\sin^2{(8x)}\,dx}

    \displaystyle = \int{\frac{1}{4}\left[\frac{1}{2} - \frac{1}{2}\cos{(16x)}\right]\,dx}

    \displaystyle = \int{\frac{1}{8}\left[1 - \cos{(16x)}\right]\,dx}.


    So I agree with your indefinite integral ( \displaystyle +C).

    Assuming you have inputted your terminals correctly, you should be correct.
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  4. #4
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    Quote Originally Posted by skeeter View Post
    your integral is fine. I evaluate it as 0.8779... change your mode to radians.
    Skeeter is correct! Change your mode to radians. This completely explains the discrepancy!
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  5. #5
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    Quote Originally Posted by SammyS View Post
    Skeeter is correct! Change your mode to radians. This completely explains the discrepancy!
    I did say "assuming you have inputted the terminals correctly"...

    I guess the saying is true, to ASSUME makes an ASS of U and ME...
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