The integral of $int_{-pi/4}^{pi/4} frac{t^4tan(t)}{2+cos(t)},dt$?












0












$begingroup$


Here is the definite integral: $$int_{-pi/4}^{pi/4} frac{t^4tan(t)}{2+cos(t)},dt$$



Apparently this evaluates to zero. I don't see any patterns at all and I have no idea how I'd evaluate it. I tried graphing this on desmos and it wasn't pretty; there are asymptotes everywhere. Shouldn't this be undefined because of the asymptotes? Even if it isn't undefined, what would the antiderivative of this possible be?



And most importantly, why does it evaluate to zero, and is there something that should've told me that?



Any help is appreciated.






calculus integration







share|cite|improve this question











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    0












    $begingroup$


    Here is the definite integral: $$int_{-pi/4}^{pi/4} frac{t^4tan(t)}{2+cos(t)},dt$$



    Apparently this evaluates to zero. I don't see any patterns at all and I have no idea how I'd evaluate it. I tried graphing this on desmos and it wasn't pretty; there are asymptotes everywhere. Shouldn't this be undefined because of the asymptotes? Even if it isn't undefined, what would the antiderivative of this possible be?



    And most importantly, why does it evaluate to zero, and is there something that should've told me that?



    Any help is appreciated.






    calculus integration







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Here is the definite integral: $$int_{-pi/4}^{pi/4} frac{t^4tan(t)}{2+cos(t)},dt$$



      Apparently this evaluates to zero. I don't see any patterns at all and I have no idea how I'd evaluate it. I tried graphing this on desmos and it wasn't pretty; there are asymptotes everywhere. Shouldn't this be undefined because of the asymptotes? Even if it isn't undefined, what would the antiderivative of this possible be?



      And most importantly, why does it evaluate to zero, and is there something that should've told me that?



      Any help is appreciated.






      calculus integration







      share|cite|improve this question











      $endgroup$




      Here is the definite integral: $$int_{-pi/4}^{pi/4} frac{t^4tan(t)}{2+cos(t)},dt$$



      Apparently this evaluates to zero. I don't see any patterns at all and I have no idea how I'd evaluate it. I tried graphing this on desmos and it wasn't pretty; there are asymptotes everywhere. Shouldn't this be undefined because of the asymptotes? Even if it isn't undefined, what would the antiderivative of this possible be?



      And most importantly, why does it evaluate to zero, and is there something that should've told me that?



      Any help is appreciated.






      calculus integration




      calculus integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 21:03









      Paul Enta

      4,81111333




      4,81111333










      asked Dec 9 '18 at 20:56









      James RonaldJames Ronald

      1207




      1207






















          4 Answers
          4






          active

          oldest

          votes


















          2












          $begingroup$

          You will find that definite integral questions with funky integrands are not generally solvable by the routine method of taking the anti-derivative, as it may be very hard to find.



          Note that here, the integrand is an odd function.



          $f(t)=frac{t^4tan t}{2+cos t}implies f(-t)=-frac{t^4tan t}{2+cos t}=-f(t)$



          Recall that for an odd function $f, int_{-a}^af=0$



          As far as asymptotes and discontinuity of $f$ is concerned, note that $f$ is only discontinuous at odd multiples of $pi/2$ and the domain of $t, (-pi/4, +pi/4)$ does not contain any points of discontinuity.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much! That was a very simply but thorough addressed all my questions, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:44



















          0












          $begingroup$

          Hint:



          $$frac{(-t)^4tan(-t)}{2+cos(-t)}=-frac{t^4tan(t)}{2+cos(t)}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes I noticed that and now the answer seems very obvious haha, thanks!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45



















          0












          $begingroup$

          $$I=int_{-a}^{a}frac{t^4tan t}{2+cos t}mathrm dt$$
          $t=-u$:
          $$I=int_{a}^{-a}frac{(-u)^4tan(-u)}{2+cos(-u)}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4(-tan u)}{2+cos u}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-int_{-a}^{a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-I$$
          $$2I=0$$
          $$I=0$$
          QED






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It's nice to see a straight up proof, didn't know you could do this using the property of even/odd, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45










          • $begingroup$
            @JamesRonald You're very welcome. It's always a good thing to try when you are unsure of how to start on an integral.
            $endgroup$
            – clathratus
            Dec 10 '18 at 20:14



















          0












          $begingroup$

          $$int_{-a}^af(t)mathrm dt=0$$
          if $f$ is an odd function. That's because odd functions are symmetric w.r.t. the origin, then if $f$ from $0$ to $a$ is defined in the first quadrant (positive area), from $-a$ to $0$ $f$ is defined in the third quadrant (negative area), thus the two areas cancel out.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That definitely makes sense, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:46










          • $begingroup$
            You are welcome!
            $endgroup$
            – Lorenzo B.
            Dec 10 '18 at 19:26











          Your Answer





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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          You will find that definite integral questions with funky integrands are not generally solvable by the routine method of taking the anti-derivative, as it may be very hard to find.



          Note that here, the integrand is an odd function.



          $f(t)=frac{t^4tan t}{2+cos t}implies f(-t)=-frac{t^4tan t}{2+cos t}=-f(t)$



          Recall that for an odd function $f, int_{-a}^af=0$



          As far as asymptotes and discontinuity of $f$ is concerned, note that $f$ is only discontinuous at odd multiples of $pi/2$ and the domain of $t, (-pi/4, +pi/4)$ does not contain any points of discontinuity.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much! That was a very simply but thorough addressed all my questions, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:44
















          2












          $begingroup$

          You will find that definite integral questions with funky integrands are not generally solvable by the routine method of taking the anti-derivative, as it may be very hard to find.



          Note that here, the integrand is an odd function.



          $f(t)=frac{t^4tan t}{2+cos t}implies f(-t)=-frac{t^4tan t}{2+cos t}=-f(t)$



          Recall that for an odd function $f, int_{-a}^af=0$



          As far as asymptotes and discontinuity of $f$ is concerned, note that $f$ is only discontinuous at odd multiples of $pi/2$ and the domain of $t, (-pi/4, +pi/4)$ does not contain any points of discontinuity.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much! That was a very simply but thorough addressed all my questions, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:44














          2












          2








          2





          $begingroup$

          You will find that definite integral questions with funky integrands are not generally solvable by the routine method of taking the anti-derivative, as it may be very hard to find.



          Note that here, the integrand is an odd function.



          $f(t)=frac{t^4tan t}{2+cos t}implies f(-t)=-frac{t^4tan t}{2+cos t}=-f(t)$



          Recall that for an odd function $f, int_{-a}^af=0$



          As far as asymptotes and discontinuity of $f$ is concerned, note that $f$ is only discontinuous at odd multiples of $pi/2$ and the domain of $t, (-pi/4, +pi/4)$ does not contain any points of discontinuity.






          share|cite|improve this answer











          $endgroup$



          You will find that definite integral questions with funky integrands are not generally solvable by the routine method of taking the anti-derivative, as it may be very hard to find.



          Note that here, the integrand is an odd function.



          $f(t)=frac{t^4tan t}{2+cos t}implies f(-t)=-frac{t^4tan t}{2+cos t}=-f(t)$



          Recall that for an odd function $f, int_{-a}^af=0$



          As far as asymptotes and discontinuity of $f$ is concerned, note that $f$ is only discontinuous at odd multiples of $pi/2$ and the domain of $t, (-pi/4, +pi/4)$ does not contain any points of discontinuity.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 11:00

























          answered Dec 9 '18 at 20:59









          Shubham JohriShubham Johri

          5,097717




          5,097717












          • $begingroup$
            Thank you so much! That was a very simply but thorough addressed all my questions, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:44


















          • $begingroup$
            Thank you so much! That was a very simply but thorough addressed all my questions, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:44
















          $begingroup$
          Thank you so much! That was a very simply but thorough addressed all my questions, thank you!
          $endgroup$
          – James Ronald
          Dec 10 '18 at 18:44




          $begingroup$
          Thank you so much! That was a very simply but thorough addressed all my questions, thank you!
          $endgroup$
          – James Ronald
          Dec 10 '18 at 18:44











          0












          $begingroup$

          Hint:



          $$frac{(-t)^4tan(-t)}{2+cos(-t)}=-frac{t^4tan(t)}{2+cos(t)}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes I noticed that and now the answer seems very obvious haha, thanks!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45
















          0












          $begingroup$

          Hint:



          $$frac{(-t)^4tan(-t)}{2+cos(-t)}=-frac{t^4tan(t)}{2+cos(t)}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes I noticed that and now the answer seems very obvious haha, thanks!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45














          0












          0








          0





          $begingroup$

          Hint:



          $$frac{(-t)^4tan(-t)}{2+cos(-t)}=-frac{t^4tan(t)}{2+cos(t)}$$






          share|cite|improve this answer









          $endgroup$



          Hint:



          $$frac{(-t)^4tan(-t)}{2+cos(-t)}=-frac{t^4tan(t)}{2+cos(t)}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 21:00









          BotondBotond

          5,6832732




          5,6832732












          • $begingroup$
            Yes I noticed that and now the answer seems very obvious haha, thanks!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45


















          • $begingroup$
            Yes I noticed that and now the answer seems very obvious haha, thanks!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45
















          $begingroup$
          Yes I noticed that and now the answer seems very obvious haha, thanks!
          $endgroup$
          – James Ronald
          Dec 10 '18 at 18:45




          $begingroup$
          Yes I noticed that and now the answer seems very obvious haha, thanks!
          $endgroup$
          – James Ronald
          Dec 10 '18 at 18:45











          0












          $begingroup$

          $$I=int_{-a}^{a}frac{t^4tan t}{2+cos t}mathrm dt$$
          $t=-u$:
          $$I=int_{a}^{-a}frac{(-u)^4tan(-u)}{2+cos(-u)}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4(-tan u)}{2+cos u}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-int_{-a}^{a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-I$$
          $$2I=0$$
          $$I=0$$
          QED






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It's nice to see a straight up proof, didn't know you could do this using the property of even/odd, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45










          • $begingroup$
            @JamesRonald You're very welcome. It's always a good thing to try when you are unsure of how to start on an integral.
            $endgroup$
            – clathratus
            Dec 10 '18 at 20:14
















          0












          $begingroup$

          $$I=int_{-a}^{a}frac{t^4tan t}{2+cos t}mathrm dt$$
          $t=-u$:
          $$I=int_{a}^{-a}frac{(-u)^4tan(-u)}{2+cos(-u)}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4(-tan u)}{2+cos u}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-int_{-a}^{a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-I$$
          $$2I=0$$
          $$I=0$$
          QED






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It's nice to see a straight up proof, didn't know you could do this using the property of even/odd, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45










          • $begingroup$
            @JamesRonald You're very welcome. It's always a good thing to try when you are unsure of how to start on an integral.
            $endgroup$
            – clathratus
            Dec 10 '18 at 20:14














          0












          0








          0





          $begingroup$

          $$I=int_{-a}^{a}frac{t^4tan t}{2+cos t}mathrm dt$$
          $t=-u$:
          $$I=int_{a}^{-a}frac{(-u)^4tan(-u)}{2+cos(-u)}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4(-tan u)}{2+cos u}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-int_{-a}^{a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-I$$
          $$2I=0$$
          $$I=0$$
          QED






          share|cite|improve this answer









          $endgroup$



          $$I=int_{-a}^{a}frac{t^4tan t}{2+cos t}mathrm dt$$
          $t=-u$:
          $$I=int_{a}^{-a}frac{(-u)^4tan(-u)}{2+cos(-u)}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4(-tan u)}{2+cos u}(-mathrm du)$$
          $$I=int_{a}^{-a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-int_{-a}^{a}frac{u^4tan u}{2+cos u}mathrm du$$
          $$I=-I$$
          $$2I=0$$
          $$I=0$$
          QED







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 21:12









          clathratusclathratus

          4,084335




          4,084335












          • $begingroup$
            It's nice to see a straight up proof, didn't know you could do this using the property of even/odd, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45










          • $begingroup$
            @JamesRonald You're very welcome. It's always a good thing to try when you are unsure of how to start on an integral.
            $endgroup$
            – clathratus
            Dec 10 '18 at 20:14


















          • $begingroup$
            It's nice to see a straight up proof, didn't know you could do this using the property of even/odd, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:45










          • $begingroup$
            @JamesRonald You're very welcome. It's always a good thing to try when you are unsure of how to start on an integral.
            $endgroup$
            – clathratus
            Dec 10 '18 at 20:14
















          $begingroup$
          It's nice to see a straight up proof, didn't know you could do this using the property of even/odd, thank you!
          $endgroup$
          – James Ronald
          Dec 10 '18 at 18:45




          $begingroup$
          It's nice to see a straight up proof, didn't know you could do this using the property of even/odd, thank you!
          $endgroup$
          – James Ronald
          Dec 10 '18 at 18:45












          $begingroup$
          @JamesRonald You're very welcome. It's always a good thing to try when you are unsure of how to start on an integral.
          $endgroup$
          – clathratus
          Dec 10 '18 at 20:14




          $begingroup$
          @JamesRonald You're very welcome. It's always a good thing to try when you are unsure of how to start on an integral.
          $endgroup$
          – clathratus
          Dec 10 '18 at 20:14











          0












          $begingroup$

          $$int_{-a}^af(t)mathrm dt=0$$
          if $f$ is an odd function. That's because odd functions are symmetric w.r.t. the origin, then if $f$ from $0$ to $a$ is defined in the first quadrant (positive area), from $-a$ to $0$ $f$ is defined in the third quadrant (negative area), thus the two areas cancel out.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That definitely makes sense, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:46










          • $begingroup$
            You are welcome!
            $endgroup$
            – Lorenzo B.
            Dec 10 '18 at 19:26
















          0












          $begingroup$

          $$int_{-a}^af(t)mathrm dt=0$$
          if $f$ is an odd function. That's because odd functions are symmetric w.r.t. the origin, then if $f$ from $0$ to $a$ is defined in the first quadrant (positive area), from $-a$ to $0$ $f$ is defined in the third quadrant (negative area), thus the two areas cancel out.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That definitely makes sense, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:46










          • $begingroup$
            You are welcome!
            $endgroup$
            – Lorenzo B.
            Dec 10 '18 at 19:26














          0












          0








          0





          $begingroup$

          $$int_{-a}^af(t)mathrm dt=0$$
          if $f$ is an odd function. That's because odd functions are symmetric w.r.t. the origin, then if $f$ from $0$ to $a$ is defined in the first quadrant (positive area), from $-a$ to $0$ $f$ is defined in the third quadrant (negative area), thus the two areas cancel out.






          share|cite|improve this answer









          $endgroup$



          $$int_{-a}^af(t)mathrm dt=0$$
          if $f$ is an odd function. That's because odd functions are symmetric w.r.t. the origin, then if $f$ from $0$ to $a$ is defined in the first quadrant (positive area), from $-a$ to $0$ $f$ is defined in the third quadrant (negative area), thus the two areas cancel out.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 21:29









          Lorenzo B.Lorenzo B.

          1,8402520




          1,8402520












          • $begingroup$
            That definitely makes sense, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:46










          • $begingroup$
            You are welcome!
            $endgroup$
            – Lorenzo B.
            Dec 10 '18 at 19:26


















          • $begingroup$
            That definitely makes sense, thank you!
            $endgroup$
            – James Ronald
            Dec 10 '18 at 18:46










          • $begingroup$
            You are welcome!
            $endgroup$
            – Lorenzo B.
            Dec 10 '18 at 19:26
















          $begingroup$
          That definitely makes sense, thank you!
          $endgroup$
          – James Ronald
          Dec 10 '18 at 18:46




          $begingroup$
          That definitely makes sense, thank you!
          $endgroup$
          – James Ronald
          Dec 10 '18 at 18:46












          $begingroup$
          You are welcome!
          $endgroup$
          – Lorenzo B.
          Dec 10 '18 at 19:26




          $begingroup$
          You are welcome!
          $endgroup$
          – Lorenzo B.
          Dec 10 '18 at 19:26


















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          Quarter-circle Tiles

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          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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