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    (Original post by Benjy100)
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    Problem 187 - THE PI CONSTRICTION - DIFFICULTY LEVEL *



    Find \int\limits_0^1 {\frac{{{x^4}{{(1 - x)}^4}}}{{1 + {x^2}}}dx}

    With the soul-shattering knowledge garnered from this wonderful integral in mind deduce that the almighty circle constant is less than \frac{{22}}{7}

    Extension work (although I know that no one in the right mind would bother with it)

    Spoiler:
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    Find \int\limits_0^1 {\frac{{{x^8}{{(1 - x)}^8}(25 + 816{x^2})}}{{3164(1 + {x^2})}}dx} AND UTILIZE THIS MAMMOTH INTEGRAL TO SHOW FURTHER THAT \pi  < \frac{{355}}{{113}}
    Whilst it's a very fun looking question with a very nice picture I'm afraid we have already had this one

    Edit: Oh and the asterisks are assumed knowledge, not difficulty
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    (Original post by Jkn)
    Whilst it's a very fun looking question with a very nice picture I'm afraid we have already had this one
    Damn I haven't been following this thread often enough. Apologies, I shall delete my post and yes I thought the apple pie was a great touch haha
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    (Original post by Benjy100)
    Damn I haven't been following this thread often enough. Apologies, I shall delete my post and yes I thought the apple pie was a great touch haha
    It's such a shame, sorry to have to point it out :lol:

    Do post another/others though
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    (Original post by Zakee)
    I was sort of joking about the computation part that I had calculated 100! x 99! x 98! x 97!... (If I did, then well, I might as well quit school now and just end up replacing the TI-84 plus).
    The highest my CASIO calculator can compute is 69! :sexface: (srs)
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    (Original post by Felix Felicis)
    The highest my CASIO calculator can compute is 69! :sexface: (srs)

    Same. Maths is just a nymphomaniac. :cool:
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    Problem 187 **/*** \int\limits_{ - \infty }^\infty  {\frac{{\cos x}}{{{x^2} + x + 1}}} dx

    Problem 188 \int\limits_0^\infty  {x{e^{ - {x^3}}}} dx
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    Right yeh..chemistry...I knew I was meant to be doing something.
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    Problem 189*

    Not hard but it's quirky :ahee:

    \displaystyle\int x e^{e^{x^{2}} +x^{2}} \ dx
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    (Original post by Benjy100)
    Problem 188 ** \int\limits_0^\infty  {x{e^{ - {x^3}}}} dx
    Solution 188

    Let u=x^3,

    \displaystyle \Rightarrow \int\limits_0^\infty {x{e^{ - {x^3}}}} dx= \frac{1}{3} \int_0^{\infty} u^{-\frac{1}{3}} e^{-u} du =\frac{1}{3} \Gamma \left(\frac{2}{3} \right) \ \square
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    (Original post by Felix Felicis)
    Solution 177

    \displaystyle\int \sqrt{ \csc x - \sin x} dx = \displaystyle\int \dfrac{ \cos x}{\sqrt{ \sin x}} dx = 2 \sqrt{ \sin x } + \mathcal{C}

    Solution 180

    \displaystyle\int \left( \ln x \cdot  \cos x + \dfrac{ \sin x}{x} \right) dx = \ln x \cdot \sin x  + \mathcal{C}

    I'll leave the rest, 183 looks like an algebraic mess :lol:
    Nice solutions bro! Why has no-one done 184? :eek: You can do it in your head it's so simple! :eek: !

    Oh and 183 isn't an algebraic mess! They are all rather trivial in fact! The intention is to see if you can do 15-25 of them in 20 minutes


    More fun ****:

    Problem 190*

    Evaluate \displaystyle\int_0^{\frac{1}{2}  } (\arcsin(x))^2 dx
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    (Original post by Jkn)
    Nice solutions bro!
    Cheers!

    (Original post by Jkn)
    Why has no-one done 184? :eek:
    (Original post by Jkn)
    You can do it in your head it's so simple! :eek: !
    Precisely this

    (Original post by Jkn)
    Oh and 183 isn't an algebraic mess! They are all rather trivial in fact! The intention is to see if you can do 15-25 of them in 20 minutes

    I know They're all quite nice, short and snappy, I could do them all now but I figured I'd let other people try them They're quite nice little integrals to do, not like some of the beasts that have been posted on here that leave you scratching your head for hours

    (Original post by Jkn)
    More fun ****:

    Problem 190*

    Evaluate \displaystyle\int_0^{\frac{1}{2}  } (\arcsin(x))^2 dx
    Solution 190

    Let x = \sin \theta :

    \displaystyle\int_{0}^{ \frac{1}{2}} \arcsin^{2} x \ dx = \displaystyle\int_{0}^{ \frac{\pi}{6}} \theta^{2} \cos \theta \ d \theta \overset{ \text{IBP}}= \dfrac{\pi^{2}}{72} + \dfrac{ \sqrt{3} \pi}{6} - 1
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    (Original post by Felix Felicis)
    Cheers!

    Precisely this

    I know They're all quite nice, short and snappy, I could do them all now but I figured I'd let other people try them They're quite nice little integrals to do, not like some of the beasts that have been posted on here that leave you scratching your head for hours

    Solution 190

    Let x = \sin \theta :

    \displaystyle\int_{0}^{ \frac{1}{2}} \arcsin^{2} x \ dx = \displaystyle\int_{0}^{ \frac{\pi}{6}} \theta^{2} \cos \theta \ d \theta \overset{ \text{IBP}}= \dfrac{\pi^{2}}{72} + \dfrac{ \sqrt{3} \pi}{6} - 1
    Mmm, they are a little too trivial for my liking though. They are more exercises than anything else I'd say and, apart from a few of the more difficult ones that have a very specific route, you tend to be able to do them with the first substitution you try :lol:

    ****, I've got 3 exams at the start of the week... I should probably gtfo this forum
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    (Original post by Jkn)
    Mmm, they are a little too trivial for my liking though. They are more exercises than anything else I'd say and, apart from a few of the more difficult ones that have a very specific route, you tend to be able to do them with the first substitution you try :lol:

    ****, I've got 3 exams at the start of the week... I should probably gtfo this forum
    True, the substitution's usually pretty obvious as well, they're just slightly harder than A-level I suppose the time pressure is the hard part :lol:
    Same, I have 2 next week and I've been procrastinating so much over half term with this stuff :lol:
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    Solution 190
    Substituting u=\arcsin(x), then x=\displaystyle\int_0^{\frac{1}{  2}  } (\arcsin(x))^2 dx = \int_0^{\frac{\pi}{6}}u^2\cos(u)  du = \left[ u^2\sin(u)+2u\cos(u)-2\sin(u) \right]_0^{\frac{\pi}{6}} = \frac{\pi^2+12\sqrt{3}\pi}{72}-1

    Also, how to format latex properly? Like, line breaks and aligned equals signs and ****

    Edit: fuuu I'm too slow
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    (Original post by Jkn)
    Nice solutions bro! Why has no-one done 184? :eek: You can do it in your head it's so simple! :eek: !

    Oh and 183 isn't an algebraic mess! They are all rather trivial in fact! The intention is to see if you can do 15-25 of them in 20 minutes


    More fun ****:

    Problem 190*

    Evaluate \displaystyle\int_0^{\frac{1}{2}  } (\arcsin(x))^2 dx
    \int^\frac{1}{2}_0 (arcsinx)^2 dx = \frac{\pi^2}{72}-\int^\frac{1}{2}_0\frac{2xarcsin  x}{\sqrt{1-x^2}} dx = \frac{\pi^2}{72}+\frac{\pi\sqrt{  3}}{6} - 1

    God typing LaTeX on a mobile is tedious.
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    Solution 187

    x\mapsto -x and adding both integrals:

    \displaystyle\int_{-\infty}^{\infty} \frac{\cos x\,dx}{x^2+x+1}=2\int_{0}^{ \infty} \frac{(x^2+1)\cos x\,dx}{x^4+x^2+1}


    \displaystyle \begin{aligned} f(t)=2\int_{0}^{ \infty} \frac{(x^2+1)\cos tx\,dx}{x^4+x^2+1} \Rightarrow \mathcal{L}\{f(t)\}&=2 \int_0^{ \infty} e^{-st}\int_{0}^{\infty} \frac{(x^2+1)\cos tx\,dx\, dt}{x^4+x^2+1}\\&=\int_0  ^{\infty}\frac{2(x^2+1)}{x^4+x^2  +1}\int_0^{  \infty}e^{-st}\cos tx\,dt\,dx\\&=\int_0^{\infty} \frac{2(x^2+1)}{x^4+x^2+1} \left(\frac{s}{x^2+s^2} \right)dx\\&=\frac{\pi}{3}\left(  \frac{2\sqrt{3}s+3}{s^2+\sqrt{3}  s+1}\right)\\&=\frac{2\pi}{\sqrt  {3}}\left(\frac{s+\frac{\sqrt{3}  }{2}}{\left(s+\frac{\sqrt{3}}{2}  \right)^2+\frac{1}{4}}\right) \Rightarrow \mathcal{L}^{-1}\left\{f(t)\right\}=\frac{2\pi  \cos \left(\frac{t}{2}\right)}{e^{ \frac{\sqrt{3}}{2}t}\sqrt{3}} \end{  aligned}

    Hence \displaystyle\int_{-\infty}^{\infty} \frac{\cos x\,dx}{x^2+x+1}=\frac{2\pi\cos \left(\frac{1}{2}\right)}{e^{ \frac{\sqrt{3}}{2}}\sqrt{3}}

    (before bananarama unleashes the swears, line 3 to 4 is done by splitting into partial fractions and completing the square etc. It's relatively messy and tedious so I omitted it).

    Problem 188**

    \displaystyle \int\limits_0^\infty  {x{e^{ - {x^3}}}} dx
    Now this is what I call mislabeling.
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    Shotgun 183.
    I may be some time typing up the tex
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    Solution 183:


    I =\displaystyle \int \dfrac{1}{\sqrt{x} (\sqrt[4]{x}+1)^{10}} dx
    Let
    u=\sqrt[4]{x} \Rightarrow dx = 4u^3 du

\Rightarrow I = 4\displaystyle \int \dfrac{u}{(u+1)^{10}} \ dx

\ p = u \Rightarrow p' =1

\ q' = \dfrac{1}{(u+1)^{10}} \Rightarrow q = -\dfrac{1}{9(u+1)^{9}}

\Rightarrow I =- \dfrac{u}{(u+1)^{9}}+ \displaystyle \int\dfrac{1}{(u+1)^{9}}+ \mathcal{C}

\Rightarrow I =- \dfrac{u}{(u+1)^{9}} -\dfrac{1}{72(u+1)^{8}}+\mathcal{  C}

\Rightarrow I =-\dfrac{\sqrt[4]{x}}{(\sqrt[4]{x}+1)^{9}} - \dfrac{1}{(\sqrt[4]{x}+1)^{8}}+\mathcal{C}

\Rightarrow I =- \dfrac{1}{18}\left( \dfrac{9\sqrt[4]{x}+1}{(\sqrt[4]{x}+1)^{9}}\right)+ \mathcal{C}
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    (Original post by Lord of the Flies)
    (before bananarama unleashes the swears, line 3 to 4 is done by splitting into partial fractions and completing the square etc. It's relatively messy and tedious so I omitted it).


    Now this is what I call mislabeling.
    :rofl: That's fair enough Explaining is fine.

    Agreed.
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    Solution 179:
    I = \displaystyle\int^{\frac{\pi}{2}  }_0 \dfrac{\sin(4x)}{\sin(x)} \ dx = \displaystyle\int^{\frac{\pi}{2}  }_0 \dfrac{2\sin(2x)\cos(2x)}{\sin(x  )} \dx

\Rightarrow I = \displaystyle\int^{\frac{\pi}{2}  }_0 4\cos(x)\cos(2x) \ dx
    Use product to sum:
    \Rightarrow I = 2\displaystyle\int^{\frac{\pi}{2  }}_0 \cos(x) + \cos(3x) \ dx 

I = \dfrac{2}{3} \left[(3\sin(x) +\sin(3x) \right]^{\frac{\pi}{2}}_0 = \frac{4}{3}
 
 
 
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