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Math Help - Trivial Integral

  1. #1
    Senior Member polymerase's Avatar
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    Trivial Integral

    Given that \int^\frac{\pi}{4}_0 \sin^2x\cos^6x\;dx=k, find the value of \int^\frac{\pi}{4}_0 \sin^4x\cos^6x\;dx in terms of k.

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  2. #2
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    Quote Originally Posted by polymerase View Post
    Given that \int^\frac{\pi}{4}_0 \sin^2x\cos^6x\;dx=k, find the value of \int^\frac{\pi}{4}_0 \sin^4x\cos^6x\;dx in terms of k.

    Thanks
    using \sin^2(x)=1-\cos^2(x) we get

    \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx-\int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx

    lets focus on the 2nd integral for a second

    \int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx let's use integration by parts with

    u=\cos^7(x) \to du=-7\sin(x)\cos^6(x) and

    dv=\sin^2(x)\cos(x)dx \to v=\frac{1}{3}\sin^3(x)

    \int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx=\frac{1}{3}\sin^3(x)\cos^7(  x)\bigg|_{0}^{\pi/4} +\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=

    \frac{1}{3} \left( \frac{1}{\sqrt{2}}\right)^3\left( \frac{1}{\sqrt{2}}\right)^7+\frac{7}{3}\int_{0}^{\  pi/4}\sin^4(x)\cos^6(x)dx=\frac{1}{96}+\frac{7}{3}\in  t_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx

    Now subbing this back into the original we get

    \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \left[ \frac{1}{96}+\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx\right]

    \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \frac{1}{96}-\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx

    Now we add \frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx to both sides of the equation to get

    \frac{10}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \frac{1}{96}

    We know that the integral on the right hand side is k so we get

    \frac{10}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=k - \frac{1}{96}

    \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\frac{3}{10} \left( k - \frac{1}{96} \right) =\frac{3}{10}k-\frac{1}{320}

    Yippy !!
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    using \sin^2(x)=1-\cos^2(x) we get

    \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx-\int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx

    lets focus on the 2nd integral for a second

    \int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx let's use integration by parts with

    u=\cos^7(x) \to du=-7\sin(x)\cos^6(x) and

    dv=\sin^2(x)\cos(x)dx \to v=\frac{1}{3}\sin^3(x)

    \int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx=\frac{1}{3}\sin^3(x)\cos^7(  x)\bigg|_{0}^{\pi/4} +\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=

    \frac{1}{3} \left( \frac{1}{\sqrt{2}}\right)^3\left( \frac{1}{\sqrt{2}}\right)^7+\frac{7}{3}\int_{0}^{\  pi/4}\sin^4(x)\cos^6(x)dx=\frac{1}{96}+\frac{7}{3}\in  t_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx

    Now subbing this back into the original we get

    \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \left[ \frac{1}{96}+\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx\right]

    \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \frac{1}{96}-\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx

    Now we add \frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx to both sides of the equation to get

    \frac{10}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \frac{1}{96}

    We know that the integral on the right hand side is k so we get

    \frac{10}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=k - \frac{1}{96}

    \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\frac{3}{10} \left( k - \frac{1}{96} \right) =\frac{3}{10}k-\frac{1}{320}

    Yippy !!
    but the answers says... \frac{3k}{10}-\frac{1}{320}
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by polymerase View Post
    but the answers says... \frac{3k}{10}-\frac{1}{320}
    Um...that's exactly the answer that TES got...

    (3/10)*k = (3k)/10
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  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    I had a typo but I fixed it. He must have saw it then
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I had a typo but I fixed it. He must have saw it then
    Oh, ok.

    why're you trying to make polymerase look bad with fractions ?!
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by Jhevon View Post
    Oh, ok.

    why're you trying to make polymerase look bad with fractions ?!
    I guess I am just EVIL at heart....

    He must have looked right after I posted becuase I noticed right after and edited it. It is really ironic. I did all of the problem but I get to the end and need to simplify \left( \frac{1}{\sqrt{2}}\right)^{10} and I guess I forgot how to use exponents .

    Brain glitch I guess.
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