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Math Help - Today's calculation of integral #15

  1. #1
    Math Engineering Student
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    Today's calculation of integral #15

    Find \int_{0}^{\pi }{x\cos ^{2}x\,dx}.

    It seems pretty trivial, but find a quick solution.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Find \int_{0}^{\pi }{x\cos ^{2}x\,dx}.

    It seems pretty trivial, but find a quick solution.
    \int_0^{\pi}\frac{x}{2}+\frac{1}{2}\int_0^{\pi}x\c  os(2x)

    Which gives \bigg[\frac{x^2}{4}+\frac{1}{2}\bigg(\frac{1}{2}x\sin(2x  )+\frac{1}{4}\cos(2x)\bigg)\bigg]\bigg|_0^{\pi}=\frac{1}{4}{\pi}^2
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  3. #3
    Math Engineering Student
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    That's not a quick solution, you omitted details that you should explain at least.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    That's not a quick solution, you omitted details that you should explain at least.
    Ok I owe it at least that much...

    \cos(x)^2=\frac{1+\cos(2x)}{2}\therefore{x\cos(x)^  2=x\bigg(\frac{1+\cos(2x)}{2}\bigg)}=\frac{x}{2}+\  frac{x\cos(2x)}{2}

    Now using integration by parts on the second and standard techniques on the first we obtain the antiderivative

    O...and just because I know you love it

    \frac{x}{2}+\frac{x\cos(2x)}{2}=\frac{x}{2}+\sum_{  n=0}^{\infty}\frac{(-1)^{n}2^{2n}x^{2n+1}}{(2n)!}

    Integrating we get

    \int{x\cos(x)^2}dx=\frac{x^2}{4}+\sum_{n=0}^{\inft  y}\frac{(-1)^{n}2^{2n}x^{2n+2}}{(2n+2)(2n)!}+C


    That was obviously a stupid joke^...me attempting to be funny
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  5. #5
    Math Engineering Student
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    Partial integration is involved. Find another solution which does not involve integration by parts.
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  6. #6
    Super Member PaulRS's Avatar
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    Let: <br />
u = \pi  - x<br />

    Then: <br />
I = \int_0^\pi  {x \cdot \cos ^2 \left( x \right) \cdot dx}  = \int_0^\pi  {\left( {\pi  - u} \right) \cdot \cos ^2 \left( u \right)du} <br /> <br />

    Summing the expressions above : <br />
I = \left( {\tfrac{\pi }<br />
{2}} \right) \cdot \int_0^\pi  {\cos ^2 \left( x \right) \cdot dx} <br /> <br />

    And: <br />
\int_0^\pi  {\cos ^2 \left( x \right) \cdot dx}  = \int_0^\pi  {\tfrac{{1 + \cos \left( {2x} \right)}}<br />
{2} \cdot dx}  = \tfrac{\pi }<br />
{2}<br />

    Thus: <br />
I = {\tfrac{\pi^2 }<br />
{4}} <br /> <br />
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  7. #7
    Math Engineering Student
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    Pretty similar I did. I defined x\to x-\frac\pi2.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    I got confused with the method shown. here's another:

    note that \int_0^\pi \cos^2 x ~dx = \int_0^\pi \sin^2 x ~dx

    Thus, \int_0^\pi x \cos^2 x ~dx = \frac 12 \left[ \int_0^\pi x \sin^2 x ~dx + \int_0^\pi x \cos^2 x~dx \right]

    = \frac 12 \left[ \int_0^\pi x ~dx\right]

    = \frac 12 \cdot \frac {x^2}2 \bigg|_0^\pi

    = \frac {\pi^2}4
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  9. #9
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Jhevon View Post
    note that \int_0^\pi \cos^2 x ~dx = \int_0^\pi \sin^2 x ~dx

    Thus, \int_0^\pi x \cos^2 x ~dx = \frac 12 \left[ \int_0^\pi x \sin^2 x ~dx + \int_0^\pi x \cos^2 x~dx \right]
    What conditions have to be respected so that one can say "if \int_a^b f(x)\,\mathrm{d}x = \int_a^b g(x)\,\mathrm{d}x then \int_a^b f(x)\cdot h(x)\,\mathrm{d}x=\int_a^b g(x)\cdot h(x)\,\mathrm{d}x" ?

    I ask this question because there are some cases in which it does not work. For example :

    \int_0^{2\pi} \sin x\,\mathrm{d}x= \int_0^{2\pi}\cos x\,\mathrm{d}x=0 but \int_0^{2\pi} \sin^2x\,\mathrm{d}x=\pi and \int_0^{2\pi} \cos x \cdot \sin x\,\mathrm{d}x=\left[\frac{sin^2x}{2}\right]_0^{2\pi}=0\neq \pi
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  10. #10
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Jhevon View Post
    I got confused with the method shown.
    Note that \cos(u)=-\cos(\pi-u), thus, when squaring on both sides, we get \cos^2(u)=\cos^2(\pi-u)
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  11. #11
    Math Engineering Student
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    Here's my solution:

    Let \lambda =\int_{0}^{\pi }{x\cos ^{2}x\,dx}. Substitute x\to x-\frac{\pi }{2}\implies \lambda =\pi \int_{0}^{\pi /2}{\sin ^{2}x\,dx}. In general \int_{0}^{\pi /2}{f(\sin x)\,dx}=\int_{0}^{\pi /2}{f(\cos x)\,dx}, hence \lambda =\frac{\pi ^{2}}{4}.
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