Doubt about splitting a definite trigonometric integral












1












$begingroup$


I was studying some already solved trigonometric integrals with absolute value.
I do not understand why this integral:
$$int_0^sqrt3|x-1|arctan(x),mathrm dx=int_0^1-(x-1)arctan(x),mathrm dx+int_1^sqrt3(x-1)arctan(x),mathrm dx$$
has to be split in two integrals, where the first one goes from $0$ to $1$. Does it have to be $1$ or can it be any number less than $sqrt3$?



Same things with this other one, why the does the first integral hast to end with $dfracπ2$?
$$int_{-pi/2}^{3pi/2}(x+1)^2|cos x|,mathrm dx=int_{-pi/2}^{pi/2}(x+1)^2|cos x|,mathrm dx-int_{pi/2}^{3pi/2}(x+1)^2|cos x|,mathrm dx.$$






calculus definite-integrals







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I was studying some already solved trigonometric integrals with absolute value.
    I do not understand why this integral:
    $$int_0^sqrt3|x-1|arctan(x),mathrm dx=int_0^1-(x-1)arctan(x),mathrm dx+int_1^sqrt3(x-1)arctan(x),mathrm dx$$
    has to be split in two integrals, where the first one goes from $0$ to $1$. Does it have to be $1$ or can it be any number less than $sqrt3$?



    Same things with this other one, why the does the first integral hast to end with $dfracπ2$?
    $$int_{-pi/2}^{3pi/2}(x+1)^2|cos x|,mathrm dx=int_{-pi/2}^{pi/2}(x+1)^2|cos x|,mathrm dx-int_{pi/2}^{3pi/2}(x+1)^2|cos x|,mathrm dx.$$






    calculus definite-integrals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I was studying some already solved trigonometric integrals with absolute value.
      I do not understand why this integral:
      $$int_0^sqrt3|x-1|arctan(x),mathrm dx=int_0^1-(x-1)arctan(x),mathrm dx+int_1^sqrt3(x-1)arctan(x),mathrm dx$$
      has to be split in two integrals, where the first one goes from $0$ to $1$. Does it have to be $1$ or can it be any number less than $sqrt3$?



      Same things with this other one, why the does the first integral hast to end with $dfracπ2$?
      $$int_{-pi/2}^{3pi/2}(x+1)^2|cos x|,mathrm dx=int_{-pi/2}^{pi/2}(x+1)^2|cos x|,mathrm dx-int_{pi/2}^{3pi/2}(x+1)^2|cos x|,mathrm dx.$$






      calculus definite-integrals







      share|cite|improve this question











      $endgroup$




      I was studying some already solved trigonometric integrals with absolute value.
      I do not understand why this integral:
      $$int_0^sqrt3|x-1|arctan(x),mathrm dx=int_0^1-(x-1)arctan(x),mathrm dx+int_1^sqrt3(x-1)arctan(x),mathrm dx$$
      has to be split in two integrals, where the first one goes from $0$ to $1$. Does it have to be $1$ or can it be any number less than $sqrt3$?



      Same things with this other one, why the does the first integral hast to end with $dfracπ2$?
      $$int_{-pi/2}^{3pi/2}(x+1)^2|cos x|,mathrm dx=int_{-pi/2}^{pi/2}(x+1)^2|cos x|,mathrm dx-int_{pi/2}^{3pi/2}(x+1)^2|cos x|,mathrm dx.$$






      calculus definite-integrals




      calculus definite-integrals






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 12:19









      Saad

      20.3k92352




      20.3k92352










      asked Jan 3 at 11:08









      El BryanEl Bryan

      417




      417






















          2 Answers
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          0












          $begingroup$

          The absolute value function $lvert x rvert$ is defined as $$lvert x rvert=begin{cases}
          x,&text{if $xgeq 0 $}\
          -x,&text{if $xlt 0 $}
          end{cases}.$$           As you can see from the definition, the function takes different values(but,always non-negative) on different intervals. So when you calculate $lvert x-1 rvert$, you have to check the intervals.
          Note that $$lvert x-1 rvert=begin{cases}
          x-1,&text{if $x-1geq 0 $}\
          -(x-1),&text{if $x-1lt 0 $}
          end{cases}.$$
          So your integral
          $$int_0^sqrt3 |x-1|operatorname{arctan}(x)mathrm{d}x$$ becomes $$int_0^1 -(x-1)operatorname{arctan}(x)mathrm{d}x + int_1^sqrt3 (x-1)operatorname{arctan}(x)mathrm{d}x$$ because $lvert x-1 rvert=-(x-1)$ on the interval $(-infty,1) $.



          The function $lvert cos x rvert$ always gives non-negative values. But
          $cos xge0$ for $-dfracpi2le xledfracpi2$
          and $cos x<0$ for $dfracpi2lt xltdfrac{3pi}2$.
          So we define $$lvert cos x rvert=begin{cases}
          cos x,&text{if $-dfracpi2le xledfracpi2$}\
          -cos x,&text{if $dfracpi2lt xltdfrac{3pi}2$}
          end{cases}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can I check that |cosx| changes on pi/2
            $endgroup$
            – El Bryan
            Jan 3 at 11:29












          • $begingroup$
            @ElBryan Do you know the quadrants on which $cos x $ is negative?
            $endgroup$
            – Thomas Shelby
            Jan 3 at 11:34










          • $begingroup$
            Now I do, but I still can't figure it out how does it work with cosx
            $endgroup$
            – El Bryan
            Jan 3 at 11:37










          • $begingroup$
            @ElBryan Can you recall the identity $cos (pi/2 +x)=-sin x $?
            $endgroup$
            – Thomas Shelby
            Jan 3 at 11:37






          • 1




            $begingroup$
            Now I get it, thanks a lot
            $endgroup$
            – El Bryan
            Jan 3 at 12:05



















          0












          $begingroup$

          The function $|x-1|$ is equal to $-(x-1)$ on $(-infty, 1]$ and is equal to $x-1$ on $[1,infty)$. So you split the integral at $1$ because that way, the two remaining integrals do not have an absolute value you would otherwise need to deal with. The split must be at $1$, otherwise you do not lose the absolute values.



          The same reason goes for the second integral you list: you split the integral there because that's where the function under the absolute value changes its sign.






          share|cite|improve this answer









          $endgroup$














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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            active

            oldest

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            0












            $begingroup$

            The absolute value function $lvert x rvert$ is defined as $$lvert x rvert=begin{cases}
            x,&text{if $xgeq 0 $}\
            -x,&text{if $xlt 0 $}
            end{cases}.$$             As you can see from the definition, the function takes different values(but,always non-negative) on different intervals. So when you calculate $lvert x-1 rvert$, you have to check the intervals.
            Note that $$lvert x-1 rvert=begin{cases}
            x-1,&text{if $x-1geq 0 $}\
            -(x-1),&text{if $x-1lt 0 $}
            end{cases}.$$
            So your integral
            $$int_0^sqrt3 |x-1|operatorname{arctan}(x)mathrm{d}x$$ becomes $$int_0^1 -(x-1)operatorname{arctan}(x)mathrm{d}x + int_1^sqrt3 (x-1)operatorname{arctan}(x)mathrm{d}x$$ because $lvert x-1 rvert=-(x-1)$ on the interval $(-infty,1) $.



            The function $lvert cos x rvert$ always gives non-negative values. But
            $cos xge0$ for $-dfracpi2le xledfracpi2$
            and $cos x<0$ for $dfracpi2lt xltdfrac{3pi}2$.
            So we define $$lvert cos x rvert=begin{cases}
            cos x,&text{if $-dfracpi2le xledfracpi2$}\
            -cos x,&text{if $dfracpi2lt xltdfrac{3pi}2$}
            end{cases}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How can I check that |cosx| changes on pi/2
              $endgroup$
              – El Bryan
              Jan 3 at 11:29












            • $begingroup$
              @ElBryan Do you know the quadrants on which $cos x $ is negative?
              $endgroup$
              – Thomas Shelby
              Jan 3 at 11:34










            • $begingroup$
              Now I do, but I still can't figure it out how does it work with cosx
              $endgroup$
              – El Bryan
              Jan 3 at 11:37










            • $begingroup$
              @ElBryan Can you recall the identity $cos (pi/2 +x)=-sin x $?
              $endgroup$
              – Thomas Shelby
              Jan 3 at 11:37






            • 1




              $begingroup$
              Now I get it, thanks a lot
              $endgroup$
              – El Bryan
              Jan 3 at 12:05
















            0












            $begingroup$

            The absolute value function $lvert x rvert$ is defined as $$lvert x rvert=begin{cases}
            x,&text{if $xgeq 0 $}\
            -x,&text{if $xlt 0 $}
            end{cases}.$$             As you can see from the definition, the function takes different values(but,always non-negative) on different intervals. So when you calculate $lvert x-1 rvert$, you have to check the intervals.
            Note that $$lvert x-1 rvert=begin{cases}
            x-1,&text{if $x-1geq 0 $}\
            -(x-1),&text{if $x-1lt 0 $}
            end{cases}.$$
            So your integral
            $$int_0^sqrt3 |x-1|operatorname{arctan}(x)mathrm{d}x$$ becomes $$int_0^1 -(x-1)operatorname{arctan}(x)mathrm{d}x + int_1^sqrt3 (x-1)operatorname{arctan}(x)mathrm{d}x$$ because $lvert x-1 rvert=-(x-1)$ on the interval $(-infty,1) $.



            The function $lvert cos x rvert$ always gives non-negative values. But
            $cos xge0$ for $-dfracpi2le xledfracpi2$
            and $cos x<0$ for $dfracpi2lt xltdfrac{3pi}2$.
            So we define $$lvert cos x rvert=begin{cases}
            cos x,&text{if $-dfracpi2le xledfracpi2$}\
            -cos x,&text{if $dfracpi2lt xltdfrac{3pi}2$}
            end{cases}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How can I check that |cosx| changes on pi/2
              $endgroup$
              – El Bryan
              Jan 3 at 11:29












            • $begingroup$
              @ElBryan Do you know the quadrants on which $cos x $ is negative?
              $endgroup$
              – Thomas Shelby
              Jan 3 at 11:34










            • $begingroup$
              Now I do, but I still can't figure it out how does it work with cosx
              $endgroup$
              – El Bryan
              Jan 3 at 11:37










            • $begingroup$
              @ElBryan Can you recall the identity $cos (pi/2 +x)=-sin x $?
              $endgroup$
              – Thomas Shelby
              Jan 3 at 11:37






            • 1




              $begingroup$
              Now I get it, thanks a lot
              $endgroup$
              – El Bryan
              Jan 3 at 12:05














            0












            0








            0





            $begingroup$

            The absolute value function $lvert x rvert$ is defined as $$lvert x rvert=begin{cases}
            x,&text{if $xgeq 0 $}\
            -x,&text{if $xlt 0 $}
            end{cases}.$$             As you can see from the definition, the function takes different values(but,always non-negative) on different intervals. So when you calculate $lvert x-1 rvert$, you have to check the intervals.
            Note that $$lvert x-1 rvert=begin{cases}
            x-1,&text{if $x-1geq 0 $}\
            -(x-1),&text{if $x-1lt 0 $}
            end{cases}.$$
            So your integral
            $$int_0^sqrt3 |x-1|operatorname{arctan}(x)mathrm{d}x$$ becomes $$int_0^1 -(x-1)operatorname{arctan}(x)mathrm{d}x + int_1^sqrt3 (x-1)operatorname{arctan}(x)mathrm{d}x$$ because $lvert x-1 rvert=-(x-1)$ on the interval $(-infty,1) $.



            The function $lvert cos x rvert$ always gives non-negative values. But
            $cos xge0$ for $-dfracpi2le xledfracpi2$
            and $cos x<0$ for $dfracpi2lt xltdfrac{3pi}2$.
            So we define $$lvert cos x rvert=begin{cases}
            cos x,&text{if $-dfracpi2le xledfracpi2$}\
            -cos x,&text{if $dfracpi2lt xltdfrac{3pi}2$}
            end{cases}.$$






            share|cite|improve this answer











            $endgroup$



            The absolute value function $lvert x rvert$ is defined as $$lvert x rvert=begin{cases}
            x,&text{if $xgeq 0 $}\
            -x,&text{if $xlt 0 $}
            end{cases}.$$             As you can see from the definition, the function takes different values(but,always non-negative) on different intervals. So when you calculate $lvert x-1 rvert$, you have to check the intervals.
            Note that $$lvert x-1 rvert=begin{cases}
            x-1,&text{if $x-1geq 0 $}\
            -(x-1),&text{if $x-1lt 0 $}
            end{cases}.$$
            So your integral
            $$int_0^sqrt3 |x-1|operatorname{arctan}(x)mathrm{d}x$$ becomes $$int_0^1 -(x-1)operatorname{arctan}(x)mathrm{d}x + int_1^sqrt3 (x-1)operatorname{arctan}(x)mathrm{d}x$$ because $lvert x-1 rvert=-(x-1)$ on the interval $(-infty,1) $.



            The function $lvert cos x rvert$ always gives non-negative values. But
            $cos xge0$ for $-dfracpi2le xledfracpi2$
            and $cos x<0$ for $dfracpi2lt xltdfrac{3pi}2$.
            So we define $$lvert cos x rvert=begin{cases}
            cos x,&text{if $-dfracpi2le xledfracpi2$}\
            -cos x,&text{if $dfracpi2lt xltdfrac{3pi}2$}
            end{cases}.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 3 at 12:47

























            answered Jan 3 at 11:25









            Thomas ShelbyThomas Shelby

            4,6012727




            4,6012727












            • $begingroup$
              How can I check that |cosx| changes on pi/2
              $endgroup$
              – El Bryan
              Jan 3 at 11:29












            • $begingroup$
              @ElBryan Do you know the quadrants on which $cos x $ is negative?
              $endgroup$
              – Thomas Shelby
              Jan 3 at 11:34










            • $begingroup$
              Now I do, but I still can't figure it out how does it work with cosx
              $endgroup$
              – El Bryan
              Jan 3 at 11:37










            • $begingroup$
              @ElBryan Can you recall the identity $cos (pi/2 +x)=-sin x $?
              $endgroup$
              – Thomas Shelby
              Jan 3 at 11:37






            • 1




              $begingroup$
              Now I get it, thanks a lot
              $endgroup$
              – El Bryan
              Jan 3 at 12:05


















            • $begingroup$
              How can I check that |cosx| changes on pi/2
              $endgroup$
              – El Bryan
              Jan 3 at 11:29












            • $begingroup$
              @ElBryan Do you know the quadrants on which $cos x $ is negative?
              $endgroup$
              – Thomas Shelby
              Jan 3 at 11:34










            • $begingroup$
              Now I do, but I still can't figure it out how does it work with cosx
              $endgroup$
              – El Bryan
              Jan 3 at 11:37










            • $begingroup$
              @ElBryan Can you recall the identity $cos (pi/2 +x)=-sin x $?
              $endgroup$
              – Thomas Shelby
              Jan 3 at 11:37






            • 1




              $begingroup$
              Now I get it, thanks a lot
              $endgroup$
              – El Bryan
              Jan 3 at 12:05
















            $begingroup$
            How can I check that |cosx| changes on pi/2
            $endgroup$
            – El Bryan
            Jan 3 at 11:29






            $begingroup$
            How can I check that |cosx| changes on pi/2
            $endgroup$
            – El Bryan
            Jan 3 at 11:29














            $begingroup$
            @ElBryan Do you know the quadrants on which $cos x $ is negative?
            $endgroup$
            – Thomas Shelby
            Jan 3 at 11:34




            $begingroup$
            @ElBryan Do you know the quadrants on which $cos x $ is negative?
            $endgroup$
            – Thomas Shelby
            Jan 3 at 11:34












            $begingroup$
            Now I do, but I still can't figure it out how does it work with cosx
            $endgroup$
            – El Bryan
            Jan 3 at 11:37




            $begingroup$
            Now I do, but I still can't figure it out how does it work with cosx
            $endgroup$
            – El Bryan
            Jan 3 at 11:37












            $begingroup$
            @ElBryan Can you recall the identity $cos (pi/2 +x)=-sin x $?
            $endgroup$
            – Thomas Shelby
            Jan 3 at 11:37




            $begingroup$
            @ElBryan Can you recall the identity $cos (pi/2 +x)=-sin x $?
            $endgroup$
            – Thomas Shelby
            Jan 3 at 11:37




            1




            1




            $begingroup$
            Now I get it, thanks a lot
            $endgroup$
            – El Bryan
            Jan 3 at 12:05




            $begingroup$
            Now I get it, thanks a lot
            $endgroup$
            – El Bryan
            Jan 3 at 12:05











            0












            $begingroup$

            The function $|x-1|$ is equal to $-(x-1)$ on $(-infty, 1]$ and is equal to $x-1$ on $[1,infty)$. So you split the integral at $1$ because that way, the two remaining integrals do not have an absolute value you would otherwise need to deal with. The split must be at $1$, otherwise you do not lose the absolute values.



            The same reason goes for the second integral you list: you split the integral there because that's where the function under the absolute value changes its sign.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The function $|x-1|$ is equal to $-(x-1)$ on $(-infty, 1]$ and is equal to $x-1$ on $[1,infty)$. So you split the integral at $1$ because that way, the two remaining integrals do not have an absolute value you would otherwise need to deal with. The split must be at $1$, otherwise you do not lose the absolute values.



              The same reason goes for the second integral you list: you split the integral there because that's where the function under the absolute value changes its sign.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The function $|x-1|$ is equal to $-(x-1)$ on $(-infty, 1]$ and is equal to $x-1$ on $[1,infty)$. So you split the integral at $1$ because that way, the two remaining integrals do not have an absolute value you would otherwise need to deal with. The split must be at $1$, otherwise you do not lose the absolute values.



                The same reason goes for the second integral you list: you split the integral there because that's where the function under the absolute value changes its sign.






                share|cite|improve this answer









                $endgroup$



                The function $|x-1|$ is equal to $-(x-1)$ on $(-infty, 1]$ and is equal to $x-1$ on $[1,infty)$. So you split the integral at $1$ because that way, the two remaining integrals do not have an absolute value you would otherwise need to deal with. The split must be at $1$, otherwise you do not lose the absolute values.



                The same reason goes for the second integral you list: you split the integral there because that's where the function under the absolute value changes its sign.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 3 at 11:13









                5xum5xum

                91.8k394161




                91.8k394161






























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