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Math Help - [SOLVED] when integrating by method of substitution for a definite integral,answer di

  1. #1
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    [SOLVED] when integrating by method of substitution for a definite integral,answer di

    I am integrating sin(squared)x cosx. With upper limit 2pi/3 and lower limit 0.
    i get Answer = 3/8.


    But when I substitute u = sinx, i get answer = 7/24.

    Why is this?

    I don't think I made any mistakes...
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  2. #2
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    Quote Originally Posted by Aerospank View Post
    I am integrating sin(squared)x cosx. With upper limit 2pi/3 and lower limit 0.
    i get Answer = 3/8.


    But when I substitute u = sinx, i get answer = 7/24.

    Why is this?

    I don't think I made any mistakes... *
    \int_0^{\frac{2\pi}{3}}\sin^2(x)\cos(x). Let z=\sin(x)\implies dz=\cos(x) dx. So then our integral is equivalent to \int_{\sin(0)}^{\sin\left(\frac{2\pi}{3}\right)}z^  2\text{ }dz=\frac{\sqrt{3}}{8}.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    \int_0^{\frac{2\pi}{3}}\sin^2(x)\cos(x). Let z=\sin(x)\implies dz=\cos(x) dx. So then our integral is equivalent to \int_{\sin(0)}^{\sin\left(\frac{2\pi}{3}\right)}z^  2\text{ }dz=\frac{\sqrt{3}}{8}.
    That is incorrect...this is the original question:
    Mathass.jpg picture by Aerospank - Photobucket

    Thanks in advance!
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  4. #4
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    Quote Originally Posted by Aerospank View Post
    That is incorrect...this is the original question:
    Mathass.jpg picture by Aerospank - Photobucket

    Thanks in advance!
    No, actually, Drexel is correct:

    \int_{0}^{sin(\frac{2 \pi }{3})} z^2 ~ dz = \int_{0}^{\frac{\sqrt{3}}{2}} z^2 ~ dz = (\frac{\sqrt{3}}{2})^3 \cdot \frac{1}{3}  = \frac{3 \cdot \sqrt{3}}{8} \cdot \frac{1}{3} = \frac{\sqrt{3}}{8}

    Perhaps next time you should bother to see that you're actually posting the correct integral (you posted \int sin^2(x)cos(x) while it is, in your page, \int cos^2(x)sin(x))...

    Also, it is not necessarily correct that cosx = \sqrt{1-sin^2x} since, depending on the angle, it could also be that cosx = -\sqrt{1-sin^2x} - the first is correct for 2(k-1)\pi \leq x \leq \frac{k}{2}\pi, ~\frac{3k}{4}\pi \leq x \leq 2k\pi while the second is correct for k\pi \leq x \leq \frac{3k}{4}\pi.
    This is apparently the mistake they wanted you to find out in that excersize (I really can't figure out if this is it since there are so many mistakes regardless).

    Also, after claiming that cos(x) = \sqrt{1-sin^2x}= \sqrt{1-u^2} they substitute it back into cos^2x\cdot sinx as u\sqrt{1-u^2} while if the first claim were true, the substitution would result in u(1-u^2).

    So there are two mistakes here. I can't tell if they are intentional or not (this text seems very low quality), but it seems as if they wanted you to note that the first fact I pointed out is what causes the different result.

    This one is another great example of why you should post the whole question and not just parts of it.
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    No, actually, Drexel is correct:

    \int_{0}^{sin(\frac{2 \pi }{3})} z^2 ~ dz = \int_{0}^{\frac{\sqrt{3}}{2}} z^2 ~ dz = (\frac{\sqrt{3}}{2})^3 \cdot \frac{1}{3}  = \frac{3 \cdot \sqrt{3}}{8} \cdot \frac{1}{3} = \frac{\sqrt{3}}{8}

    Perhaps next time you should bother to see that you're actually posting the correct integral (you posted \int sin^2(x)cos(x) while it is, in your page, \int cos^2(x)sin(x))...

    Also, it is not necessarily correct that cosx = \sqrt{1-sin^2x} since, depending on the angle, it could also be that cosx = -\sqrt{1-sin^2x} - the first is correct for 2(k-1)\pi \leq x \leq \frac{k}{2}\pi, ~\frac{3k}{4}\pi \leq x \leq 2k\pi while the second is correct for k\pi \leq x \leq \frac{3k}{4}\pi.
    This is apparently the mistake they wanted you to find out in that excersize (I really can't figure out if this is it since there are so many mistakes regardless).

    Also, after claiming that cos(x) = \sqrt{1-sin^2x}= \sqrt{1-u^2} they substitute it back into cos^2x\cdot sinx as u\sqrt{1-u^2} while if the first claim were true, the substitution would result in u(1-u^2).

    So there are two mistakes here. I can't tell if they are intentional or not (this text seems very low quality), but it seems as if they wanted you to note that the first fact I pointed out is what causes the different result.

    This one is another great example of why you should post the whole question and not just parts of it.
    Thank you so much. I will be careful in posting the correct question next time.

    To Drexel: Sorry..and thanks for your kindness to help.
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  6. #6
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    Quote Originally Posted by Defunkt View Post

    Also, after claiming that cos(x) = \sqrt{1-sin^2x}= \sqrt{1-u^2} they substitute it back into cos^2x\cdot sinx as u\sqrt{1-u^2} while if the first claim were true, the substitution would result in u(1-u^2).
    ah, the result is u\sqrt{1-u^2} because if you think of it, du = cosx dx. therefore dx = du/cosx. When substituting only this 'dx' expression in the integral at first, one of the cosx cancels out.
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  7. #7
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    That is correct, my bad. However, the mistake is where they used cosx=\sqrt{1-sin^2x}.

    As you can see here, using cosx = -\sqrt{1-sin^2x} in the second quadrant gives us the correct result.
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