Integral $intfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} mathrm d x$












3












$begingroup$



Integrate $displaystyleintdfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} mathrm d x$




I tried dividing by $cos^2(x)$ and then substituting $tan(x)=t$.






real-analysis integration derivatives indefinite-integrals







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  • $begingroup$
    Please use Mathjax to typeset your equations.
    $endgroup$
    – Shubham Johri
    Feb 2 at 6:40
















3












$begingroup$



Integrate $displaystyleintdfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} mathrm d x$




I tried dividing by $cos^2(x)$ and then substituting $tan(x)=t$.






real-analysis integration derivatives indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please use Mathjax to typeset your equations.
    $endgroup$
    – Shubham Johri
    Feb 2 at 6:40














3












3








3


0



$begingroup$



Integrate $displaystyleintdfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} mathrm d x$




I tried dividing by $cos^2(x)$ and then substituting $tan(x)=t$.






real-analysis integration derivatives indefinite-integrals







share|cite|improve this question











$endgroup$





Integrate $displaystyleintdfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} mathrm d x$




I tried dividing by $cos^2(x)$ and then substituting $tan(x)=t$.






real-analysis integration derivatives indefinite-integrals




real-analysis integration derivatives indefinite-integrals






share|cite|improve this question















share|cite|improve this question













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edited Feb 3 at 3:33









Thomas Shelby

4,7382727




4,7382727










asked Feb 2 at 6:35









Meet ShahMeet Shah

575




575












  • $begingroup$
    Please use Mathjax to typeset your equations.
    $endgroup$
    – Shubham Johri
    Feb 2 at 6:40


















  • $begingroup$
    Please use Mathjax to typeset your equations.
    $endgroup$
    – Shubham Johri
    Feb 2 at 6:40
















$begingroup$
Please use Mathjax to typeset your equations.
$endgroup$
– Shubham Johri
Feb 2 at 6:40




$begingroup$
Please use Mathjax to typeset your equations.
$endgroup$
– Shubham Johri
Feb 2 at 6:40










5 Answers
5






active

oldest

votes


















2












$begingroup$

After substituting $cos(2x)=cos^2x-sin^2x$ and $sin(2x)=2sin xcos x$ and doing a bit of algebra, we find that



$$2xcos(2x)+(x^2-1)sin(2x)=2(xsin x+cos x)(xcos x-sin x)$$



and thus



$$begin{align}{2x^2over 2xcos(2x)+(x^2-1)sin(2x)}
&={x^2over(xsin x+cos x)(xcos x-sin x)}\
&={xcos xover xsin x+cos x}+{xsin xover xcos x-sin x}end{align}$$



The substitution $u=xsin x+cos x$, $du=xcos x,dx$ tells us



$$int{xcos xover xsin x+cos x},dx=int{duover u}=ln|u|+C=ln|xsin x+cos x|+C$$



while the substitution $u=xcos x-sin x$, $du=-xsin x$ tells us



$$int{xsin xover xcos x-sin x},dx=-int{duover u}=-ln|u|+C = -ln|xcos x-sin x|+C$$



and thus



$$int{2x^2over 2xcos(2x)+(x^2-1)sin(2x)},dx=lnleft|xsin x+cos xover xcos x-sin xright|+C$$



Remark: The key here was to get split the integral into two pieces, each of which is easily handled with a simple substitution. The two pieces' substitutions are similar, but not the same. It was not obvious (to me, at least) that the double-angle formulae would lead to a nice factorization, nor that "partial fractions" on the factored denominator would produce such nice results, but the individual steps seemed natural to consider.






share|cite|improve this answer











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  • $begingroup$
    Thanks , that's exactly what I was looking for.
    $endgroup$
    – Meet Shah
    Feb 8 at 6:18





















3












$begingroup$

Here's a solution using @MichaelRozenberg's hint. Solution requires some careful observation and without Michael's hint, I wouldn't have noticed it. Still, it provides a quick answer.



Using double-angle formula, we get



$$dfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)}=dfrac{x^2}{(xcos x-sin x)(xsin x+cos x)}.$$
Dividing the numerator and denominator by $cos^2x $ gives, $$dfrac{x^2sec^2x}{(x-tan x)(xtan x+1)}.$$
Now observe that $$left(dfrac{x-tan x}{xtan x+1}right)^{'}=dfrac{-x^2sec^2x}{(xtan x+1)^2}.$$
So multiply the numerator and denominator by $xtan x+1$ and take $dfrac{x-tan x}{xtan x+1}=t. $






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$endgroup$





















    3












    $begingroup$

    The hint:



    Compute
    $$left(lndfrac{x-tan x}{xtan x+1}right)^{'}.$$






    share|cite|improve this answer











    $endgroup$









    • 4




      $begingroup$
      Thats working it backwards
      $endgroup$
      – Meet Shah
      Feb 2 at 6:54












    • $begingroup$
      @Meet Shah Indeed, because we need to calculate the integral.
      $endgroup$
      – Michael Rozenberg
      Feb 2 at 6:56






    • 5




      $begingroup$
      I wouldn't know the answer in tests. You wouldn't have seen that its a differntial of that function if i didn't mention the answer.
      $endgroup$
      – Meet Shah
      Feb 2 at 7:01






    • 3




      $begingroup$
      @MichaelRozenberg The OP wants to ask: how did you manage to find that expression in the first place. I have no doubt the OP can verify compute that by reverse engineering; they are interested in the steps to reach the antiderivative. Producing a result without working is extremely unhelpful.
      $endgroup$
      – TheSimpliFire
      Feb 2 at 13:40








    • 1




      $begingroup$
      @MichaelRozenberg This is still at best a comment. A good answer should explain, not state.
      $endgroup$
      – TheSimpliFire
      Feb 2 at 16:32



















    2












    $begingroup$

    $$2xcos (2x)+(x^2-1)sin 2x=(x^2+1)bigg[frac{2x}{x^2+1}cos 2x+frac{x^2-1}{x^2+1}sin 2xbigg]$$



    $$=(x^2+1)cosbigg(2x-2alphabigg)$$



    $displaystyle sin(2alpha)=frac{x^2-1}{x^2+1}$ and $displaystyle cos(2alpha)=frac{2x}{x^2+1}$



    integration $$=intsecbigg(2x-tan^{-1}frac{x^2-1}{2x}bigg)frac{2x^2}{x^2+1}dx$$



    put $displaystyle 2x-tan^{-1}bigg(frac{x^2-1}{2x}bigg)=t$



    And $displaystyle frac{2x^2}{x^2+1}dx=dt$



    Integral $$=int sec(t)dt=lnbigg|sec (t)+tan (t)bigg|+C$$






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      Let $arctan x=yimplies x=tan y,x^2-1=dfrac{sin^2y}{cos^2y}-1=-(1+x^2)cos2y$



      $$2xcos(2x)+(x^2-1)sin2x=(1+x^2)sin2(arctan x-x)$$



      $$I=displaystyleintdfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} =intdfrac{2x^2}{(1+x^2)sin2(arctan x-x)}$$



      Now set $u=arctan x-x,du=-dfrac{x^2 dx}{1+x^2}$



      $$I=-2intdfrac{du}{sin2u}=?$$



      Use this






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        $$-2intcsc2u du=-ln(csc2u+cot2u)+K=ln(csc2u-cot2u)+K=ln(tan u)+K$$
        $endgroup$
        – lab bhattacharjee
        Feb 2 at 14:56












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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      After substituting $cos(2x)=cos^2x-sin^2x$ and $sin(2x)=2sin xcos x$ and doing a bit of algebra, we find that



      $$2xcos(2x)+(x^2-1)sin(2x)=2(xsin x+cos x)(xcos x-sin x)$$



      and thus



      $$begin{align}{2x^2over 2xcos(2x)+(x^2-1)sin(2x)}
      &={x^2over(xsin x+cos x)(xcos x-sin x)}\
      &={xcos xover xsin x+cos x}+{xsin xover xcos x-sin x}end{align}$$



      The substitution $u=xsin x+cos x$, $du=xcos x,dx$ tells us



      $$int{xcos xover xsin x+cos x},dx=int{duover u}=ln|u|+C=ln|xsin x+cos x|+C$$



      while the substitution $u=xcos x-sin x$, $du=-xsin x$ tells us



      $$int{xsin xover xcos x-sin x},dx=-int{duover u}=-ln|u|+C = -ln|xcos x-sin x|+C$$



      and thus



      $$int{2x^2over 2xcos(2x)+(x^2-1)sin(2x)},dx=lnleft|xsin x+cos xover xcos x-sin xright|+C$$



      Remark: The key here was to get split the integral into two pieces, each of which is easily handled with a simple substitution. The two pieces' substitutions are similar, but not the same. It was not obvious (to me, at least) that the double-angle formulae would lead to a nice factorization, nor that "partial fractions" on the factored denominator would produce such nice results, but the individual steps seemed natural to consider.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thanks , that's exactly what I was looking for.
        $endgroup$
        – Meet Shah
        Feb 8 at 6:18


















      2












      $begingroup$

      After substituting $cos(2x)=cos^2x-sin^2x$ and $sin(2x)=2sin xcos x$ and doing a bit of algebra, we find that



      $$2xcos(2x)+(x^2-1)sin(2x)=2(xsin x+cos x)(xcos x-sin x)$$



      and thus



      $$begin{align}{2x^2over 2xcos(2x)+(x^2-1)sin(2x)}
      &={x^2over(xsin x+cos x)(xcos x-sin x)}\
      &={xcos xover xsin x+cos x}+{xsin xover xcos x-sin x}end{align}$$



      The substitution $u=xsin x+cos x$, $du=xcos x,dx$ tells us



      $$int{xcos xover xsin x+cos x},dx=int{duover u}=ln|u|+C=ln|xsin x+cos x|+C$$



      while the substitution $u=xcos x-sin x$, $du=-xsin x$ tells us



      $$int{xsin xover xcos x-sin x},dx=-int{duover u}=-ln|u|+C = -ln|xcos x-sin x|+C$$



      and thus



      $$int{2x^2over 2xcos(2x)+(x^2-1)sin(2x)},dx=lnleft|xsin x+cos xover xcos x-sin xright|+C$$



      Remark: The key here was to get split the integral into two pieces, each of which is easily handled with a simple substitution. The two pieces' substitutions are similar, but not the same. It was not obvious (to me, at least) that the double-angle formulae would lead to a nice factorization, nor that "partial fractions" on the factored denominator would produce such nice results, but the individual steps seemed natural to consider.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thanks , that's exactly what I was looking for.
        $endgroup$
        – Meet Shah
        Feb 8 at 6:18
















      2












      2








      2





      $begingroup$

      After substituting $cos(2x)=cos^2x-sin^2x$ and $sin(2x)=2sin xcos x$ and doing a bit of algebra, we find that



      $$2xcos(2x)+(x^2-1)sin(2x)=2(xsin x+cos x)(xcos x-sin x)$$



      and thus



      $$begin{align}{2x^2over 2xcos(2x)+(x^2-1)sin(2x)}
      &={x^2over(xsin x+cos x)(xcos x-sin x)}\
      &={xcos xover xsin x+cos x}+{xsin xover xcos x-sin x}end{align}$$



      The substitution $u=xsin x+cos x$, $du=xcos x,dx$ tells us



      $$int{xcos xover xsin x+cos x},dx=int{duover u}=ln|u|+C=ln|xsin x+cos x|+C$$



      while the substitution $u=xcos x-sin x$, $du=-xsin x$ tells us



      $$int{xsin xover xcos x-sin x},dx=-int{duover u}=-ln|u|+C = -ln|xcos x-sin x|+C$$



      and thus



      $$int{2x^2over 2xcos(2x)+(x^2-1)sin(2x)},dx=lnleft|xsin x+cos xover xcos x-sin xright|+C$$



      Remark: The key here was to get split the integral into two pieces, each of which is easily handled with a simple substitution. The two pieces' substitutions are similar, but not the same. It was not obvious (to me, at least) that the double-angle formulae would lead to a nice factorization, nor that "partial fractions" on the factored denominator would produce such nice results, but the individual steps seemed natural to consider.






      share|cite|improve this answer











      $endgroup$



      After substituting $cos(2x)=cos^2x-sin^2x$ and $sin(2x)=2sin xcos x$ and doing a bit of algebra, we find that



      $$2xcos(2x)+(x^2-1)sin(2x)=2(xsin x+cos x)(xcos x-sin x)$$



      and thus



      $$begin{align}{2x^2over 2xcos(2x)+(x^2-1)sin(2x)}
      &={x^2over(xsin x+cos x)(xcos x-sin x)}\
      &={xcos xover xsin x+cos x}+{xsin xover xcos x-sin x}end{align}$$



      The substitution $u=xsin x+cos x$, $du=xcos x,dx$ tells us



      $$int{xcos xover xsin x+cos x},dx=int{duover u}=ln|u|+C=ln|xsin x+cos x|+C$$



      while the substitution $u=xcos x-sin x$, $du=-xsin x$ tells us



      $$int{xsin xover xcos x-sin x},dx=-int{duover u}=-ln|u|+C = -ln|xcos x-sin x|+C$$



      and thus



      $$int{2x^2over 2xcos(2x)+(x^2-1)sin(2x)},dx=lnleft|xsin x+cos xover xcos x-sin xright|+C$$



      Remark: The key here was to get split the integral into two pieces, each of which is easily handled with a simple substitution. The two pieces' substitutions are similar, but not the same. It was not obvious (to me, at least) that the double-angle formulae would lead to a nice factorization, nor that "partial fractions" on the factored denominator would produce such nice results, but the individual steps seemed natural to consider.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Feb 2 at 16:56

























      answered Feb 2 at 16:46









      Barry CipraBarry Cipra

      60.7k655129




      60.7k655129












      • $begingroup$
        Thanks , that's exactly what I was looking for.
        $endgroup$
        – Meet Shah
        Feb 8 at 6:18




















      • $begingroup$
        Thanks , that's exactly what I was looking for.
        $endgroup$
        – Meet Shah
        Feb 8 at 6:18


















      $begingroup$
      Thanks , that's exactly what I was looking for.
      $endgroup$
      – Meet Shah
      Feb 8 at 6:18






      $begingroup$
      Thanks , that's exactly what I was looking for.
      $endgroup$
      – Meet Shah
      Feb 8 at 6:18













      3












      $begingroup$

      Here's a solution using @MichaelRozenberg's hint. Solution requires some careful observation and without Michael's hint, I wouldn't have noticed it. Still, it provides a quick answer.



      Using double-angle formula, we get



      $$dfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)}=dfrac{x^2}{(xcos x-sin x)(xsin x+cos x)}.$$
      Dividing the numerator and denominator by $cos^2x $ gives, $$dfrac{x^2sec^2x}{(x-tan x)(xtan x+1)}.$$
      Now observe that $$left(dfrac{x-tan x}{xtan x+1}right)^{'}=dfrac{-x^2sec^2x}{(xtan x+1)^2}.$$
      So multiply the numerator and denominator by $xtan x+1$ and take $dfrac{x-tan x}{xtan x+1}=t. $






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Here's a solution using @MichaelRozenberg's hint. Solution requires some careful observation and without Michael's hint, I wouldn't have noticed it. Still, it provides a quick answer.



        Using double-angle formula, we get



        $$dfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)}=dfrac{x^2}{(xcos x-sin x)(xsin x+cos x)}.$$
        Dividing the numerator and denominator by $cos^2x $ gives, $$dfrac{x^2sec^2x}{(x-tan x)(xtan x+1)}.$$
        Now observe that $$left(dfrac{x-tan x}{xtan x+1}right)^{'}=dfrac{-x^2sec^2x}{(xtan x+1)^2}.$$
        So multiply the numerator and denominator by $xtan x+1$ and take $dfrac{x-tan x}{xtan x+1}=t. $






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Here's a solution using @MichaelRozenberg's hint. Solution requires some careful observation and without Michael's hint, I wouldn't have noticed it. Still, it provides a quick answer.



          Using double-angle formula, we get



          $$dfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)}=dfrac{x^2}{(xcos x-sin x)(xsin x+cos x)}.$$
          Dividing the numerator and denominator by $cos^2x $ gives, $$dfrac{x^2sec^2x}{(x-tan x)(xtan x+1)}.$$
          Now observe that $$left(dfrac{x-tan x}{xtan x+1}right)^{'}=dfrac{-x^2sec^2x}{(xtan x+1)^2}.$$
          So multiply the numerator and denominator by $xtan x+1$ and take $dfrac{x-tan x}{xtan x+1}=t. $






          share|cite|improve this answer









          $endgroup$



          Here's a solution using @MichaelRozenberg's hint. Solution requires some careful observation and without Michael's hint, I wouldn't have noticed it. Still, it provides a quick answer.



          Using double-angle formula, we get



          $$dfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)}=dfrac{x^2}{(xcos x-sin x)(xsin x+cos x)}.$$
          Dividing the numerator and denominator by $cos^2x $ gives, $$dfrac{x^2sec^2x}{(x-tan x)(xtan x+1)}.$$
          Now observe that $$left(dfrac{x-tan x}{xtan x+1}right)^{'}=dfrac{-x^2sec^2x}{(xtan x+1)^2}.$$
          So multiply the numerator and denominator by $xtan x+1$ and take $dfrac{x-tan x}{xtan x+1}=t. $







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 2 at 15:04









          Thomas ShelbyThomas Shelby

          4,7382727




          4,7382727























              3












              $begingroup$

              The hint:



              Compute
              $$left(lndfrac{x-tan x}{xtan x+1}right)^{'}.$$






              share|cite|improve this answer











              $endgroup$









              • 4




                $begingroup$
                Thats working it backwards
                $endgroup$
                – Meet Shah
                Feb 2 at 6:54












              • $begingroup$
                @Meet Shah Indeed, because we need to calculate the integral.
                $endgroup$
                – Michael Rozenberg
                Feb 2 at 6:56






              • 5




                $begingroup$
                I wouldn't know the answer in tests. You wouldn't have seen that its a differntial of that function if i didn't mention the answer.
                $endgroup$
                – Meet Shah
                Feb 2 at 7:01






              • 3




                $begingroup$
                @MichaelRozenberg The OP wants to ask: how did you manage to find that expression in the first place. I have no doubt the OP can verify compute that by reverse engineering; they are interested in the steps to reach the antiderivative. Producing a result without working is extremely unhelpful.
                $endgroup$
                – TheSimpliFire
                Feb 2 at 13:40








              • 1




                $begingroup$
                @MichaelRozenberg This is still at best a comment. A good answer should explain, not state.
                $endgroup$
                – TheSimpliFire
                Feb 2 at 16:32
















              3












              $begingroup$

              The hint:



              Compute
              $$left(lndfrac{x-tan x}{xtan x+1}right)^{'}.$$






              share|cite|improve this answer











              $endgroup$









              • 4




                $begingroup$
                Thats working it backwards
                $endgroup$
                – Meet Shah
                Feb 2 at 6:54












              • $begingroup$
                @Meet Shah Indeed, because we need to calculate the integral.
                $endgroup$
                – Michael Rozenberg
                Feb 2 at 6:56






              • 5




                $begingroup$
                I wouldn't know the answer in tests. You wouldn't have seen that its a differntial of that function if i didn't mention the answer.
                $endgroup$
                – Meet Shah
                Feb 2 at 7:01






              • 3




                $begingroup$
                @MichaelRozenberg The OP wants to ask: how did you manage to find that expression in the first place. I have no doubt the OP can verify compute that by reverse engineering; they are interested in the steps to reach the antiderivative. Producing a result without working is extremely unhelpful.
                $endgroup$
                – TheSimpliFire
                Feb 2 at 13:40








              • 1




                $begingroup$
                @MichaelRozenberg This is still at best a comment. A good answer should explain, not state.
                $endgroup$
                – TheSimpliFire
                Feb 2 at 16:32














              3












              3








              3





              $begingroup$

              The hint:



              Compute
              $$left(lndfrac{x-tan x}{xtan x+1}right)^{'}.$$






              share|cite|improve this answer











              $endgroup$



              The hint:



              Compute
              $$left(lndfrac{x-tan x}{xtan x+1}right)^{'}.$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Feb 2 at 15:31









              Thomas Shelby

              4,7382727




              4,7382727










              answered Feb 2 at 6:54









              Michael RozenbergMichael Rozenberg

              110k1896201




              110k1896201








              • 4




                $begingroup$
                Thats working it backwards
                $endgroup$
                – Meet Shah
                Feb 2 at 6:54












              • $begingroup$
                @Meet Shah Indeed, because we need to calculate the integral.
                $endgroup$
                – Michael Rozenberg
                Feb 2 at 6:56






              • 5




                $begingroup$
                I wouldn't know the answer in tests. You wouldn't have seen that its a differntial of that function if i didn't mention the answer.
                $endgroup$
                – Meet Shah
                Feb 2 at 7:01






              • 3




                $begingroup$
                @MichaelRozenberg The OP wants to ask: how did you manage to find that expression in the first place. I have no doubt the OP can verify compute that by reverse engineering; they are interested in the steps to reach the antiderivative. Producing a result without working is extremely unhelpful.
                $endgroup$
                – TheSimpliFire
                Feb 2 at 13:40








              • 1




                $begingroup$
                @MichaelRozenberg This is still at best a comment. A good answer should explain, not state.
                $endgroup$
                – TheSimpliFire
                Feb 2 at 16:32














              • 4




                $begingroup$
                Thats working it backwards
                $endgroup$
                – Meet Shah
                Feb 2 at 6:54












              • $begingroup$
                @Meet Shah Indeed, because we need to calculate the integral.
                $endgroup$
                – Michael Rozenberg
                Feb 2 at 6:56






              • 5




                $begingroup$
                I wouldn't know the answer in tests. You wouldn't have seen that its a differntial of that function if i didn't mention the answer.
                $endgroup$
                – Meet Shah
                Feb 2 at 7:01






              • 3




                $begingroup$
                @MichaelRozenberg The OP wants to ask: how did you manage to find that expression in the first place. I have no doubt the OP can verify compute that by reverse engineering; they are interested in the steps to reach the antiderivative. Producing a result without working is extremely unhelpful.
                $endgroup$
                – TheSimpliFire
                Feb 2 at 13:40








              • 1




                $begingroup$
                @MichaelRozenberg This is still at best a comment. A good answer should explain, not state.
                $endgroup$
                – TheSimpliFire
                Feb 2 at 16:32








              4




              4




              $begingroup$
              Thats working it backwards
              $endgroup$
              – Meet Shah
              Feb 2 at 6:54






              $begingroup$
              Thats working it backwards
              $endgroup$
              – Meet Shah
              Feb 2 at 6:54














              $begingroup$
              @Meet Shah Indeed, because we need to calculate the integral.
              $endgroup$
              – Michael Rozenberg
              Feb 2 at 6:56




              $begingroup$
              @Meet Shah Indeed, because we need to calculate the integral.
              $endgroup$
              – Michael Rozenberg
              Feb 2 at 6:56




              5




              5




              $begingroup$
              I wouldn't know the answer in tests. You wouldn't have seen that its a differntial of that function if i didn't mention the answer.
              $endgroup$
              – Meet Shah
              Feb 2 at 7:01




              $begingroup$
              I wouldn't know the answer in tests. You wouldn't have seen that its a differntial of that function if i didn't mention the answer.
              $endgroup$
              – Meet Shah
              Feb 2 at 7:01




              3




              3




              $begingroup$
              @MichaelRozenberg The OP wants to ask: how did you manage to find that expression in the first place. I have no doubt the OP can verify compute that by reverse engineering; they are interested in the steps to reach the antiderivative. Producing a result without working is extremely unhelpful.
              $endgroup$
              – TheSimpliFire
              Feb 2 at 13:40






              $begingroup$
              @MichaelRozenberg The OP wants to ask: how did you manage to find that expression in the first place. I have no doubt the OP can verify compute that by reverse engineering; they are interested in the steps to reach the antiderivative. Producing a result without working is extremely unhelpful.
              $endgroup$
              – TheSimpliFire
              Feb 2 at 13:40






              1




              1




              $begingroup$
              @MichaelRozenberg This is still at best a comment. A good answer should explain, not state.
              $endgroup$
              – TheSimpliFire
              Feb 2 at 16:32




              $begingroup$
              @MichaelRozenberg This is still at best a comment. A good answer should explain, not state.
              $endgroup$
              – TheSimpliFire
              Feb 2 at 16:32











              2












              $begingroup$

              $$2xcos (2x)+(x^2-1)sin 2x=(x^2+1)bigg[frac{2x}{x^2+1}cos 2x+frac{x^2-1}{x^2+1}sin 2xbigg]$$



              $$=(x^2+1)cosbigg(2x-2alphabigg)$$



              $displaystyle sin(2alpha)=frac{x^2-1}{x^2+1}$ and $displaystyle cos(2alpha)=frac{2x}{x^2+1}$



              integration $$=intsecbigg(2x-tan^{-1}frac{x^2-1}{2x}bigg)frac{2x^2}{x^2+1}dx$$



              put $displaystyle 2x-tan^{-1}bigg(frac{x^2-1}{2x}bigg)=t$



              And $displaystyle frac{2x^2}{x^2+1}dx=dt$



              Integral $$=int sec(t)dt=lnbigg|sec (t)+tan (t)bigg|+C$$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                $$2xcos (2x)+(x^2-1)sin 2x=(x^2+1)bigg[frac{2x}{x^2+1}cos 2x+frac{x^2-1}{x^2+1}sin 2xbigg]$$



                $$=(x^2+1)cosbigg(2x-2alphabigg)$$



                $displaystyle sin(2alpha)=frac{x^2-1}{x^2+1}$ and $displaystyle cos(2alpha)=frac{2x}{x^2+1}$



                integration $$=intsecbigg(2x-tan^{-1}frac{x^2-1}{2x}bigg)frac{2x^2}{x^2+1}dx$$



                put $displaystyle 2x-tan^{-1}bigg(frac{x^2-1}{2x}bigg)=t$



                And $displaystyle frac{2x^2}{x^2+1}dx=dt$



                Integral $$=int sec(t)dt=lnbigg|sec (t)+tan (t)bigg|+C$$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $$2xcos (2x)+(x^2-1)sin 2x=(x^2+1)bigg[frac{2x}{x^2+1}cos 2x+frac{x^2-1}{x^2+1}sin 2xbigg]$$



                  $$=(x^2+1)cosbigg(2x-2alphabigg)$$



                  $displaystyle sin(2alpha)=frac{x^2-1}{x^2+1}$ and $displaystyle cos(2alpha)=frac{2x}{x^2+1}$



                  integration $$=intsecbigg(2x-tan^{-1}frac{x^2-1}{2x}bigg)frac{2x^2}{x^2+1}dx$$



                  put $displaystyle 2x-tan^{-1}bigg(frac{x^2-1}{2x}bigg)=t$



                  And $displaystyle frac{2x^2}{x^2+1}dx=dt$



                  Integral $$=int sec(t)dt=lnbigg|sec (t)+tan (t)bigg|+C$$






                  share|cite|improve this answer











                  $endgroup$



                  $$2xcos (2x)+(x^2-1)sin 2x=(x^2+1)bigg[frac{2x}{x^2+1}cos 2x+frac{x^2-1}{x^2+1}sin 2xbigg]$$



                  $$=(x^2+1)cosbigg(2x-2alphabigg)$$



                  $displaystyle sin(2alpha)=frac{x^2-1}{x^2+1}$ and $displaystyle cos(2alpha)=frac{2x}{x^2+1}$



                  integration $$=intsecbigg(2x-tan^{-1}frac{x^2-1}{2x}bigg)frac{2x^2}{x^2+1}dx$$



                  put $displaystyle 2x-tan^{-1}bigg(frac{x^2-1}{2x}bigg)=t$



                  And $displaystyle frac{2x^2}{x^2+1}dx=dt$



                  Integral $$=int sec(t)dt=lnbigg|sec (t)+tan (t)bigg|+C$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 2 at 10:04

























                  answered Feb 2 at 7:01









                  jackyjacky

                  1,349816




                  1,349816























                      2












                      $begingroup$

                      Let $arctan x=yimplies x=tan y,x^2-1=dfrac{sin^2y}{cos^2y}-1=-(1+x^2)cos2y$



                      $$2xcos(2x)+(x^2-1)sin2x=(1+x^2)sin2(arctan x-x)$$



                      $$I=displaystyleintdfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} =intdfrac{2x^2}{(1+x^2)sin2(arctan x-x)}$$



                      Now set $u=arctan x-x,du=-dfrac{x^2 dx}{1+x^2}$



                      $$I=-2intdfrac{du}{sin2u}=?$$



                      Use this






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        $$-2intcsc2u du=-ln(csc2u+cot2u)+K=ln(csc2u-cot2u)+K=ln(tan u)+K$$
                        $endgroup$
                        – lab bhattacharjee
                        Feb 2 at 14:56
















                      2












                      $begingroup$

                      Let $arctan x=yimplies x=tan y,x^2-1=dfrac{sin^2y}{cos^2y}-1=-(1+x^2)cos2y$



                      $$2xcos(2x)+(x^2-1)sin2x=(1+x^2)sin2(arctan x-x)$$



                      $$I=displaystyleintdfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} =intdfrac{2x^2}{(1+x^2)sin2(arctan x-x)}$$



                      Now set $u=arctan x-x,du=-dfrac{x^2 dx}{1+x^2}$



                      $$I=-2intdfrac{du}{sin2u}=?$$



                      Use this






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        $$-2intcsc2u du=-ln(csc2u+cot2u)+K=ln(csc2u-cot2u)+K=ln(tan u)+K$$
                        $endgroup$
                        – lab bhattacharjee
                        Feb 2 at 14:56














                      2












                      2








                      2





                      $begingroup$

                      Let $arctan x=yimplies x=tan y,x^2-1=dfrac{sin^2y}{cos^2y}-1=-(1+x^2)cos2y$



                      $$2xcos(2x)+(x^2-1)sin2x=(1+x^2)sin2(arctan x-x)$$



                      $$I=displaystyleintdfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} =intdfrac{2x^2}{(1+x^2)sin2(arctan x-x)}$$



                      Now set $u=arctan x-x,du=-dfrac{x^2 dx}{1+x^2}$



                      $$I=-2intdfrac{du}{sin2u}=?$$



                      Use this






                      share|cite|improve this answer









                      $endgroup$



                      Let $arctan x=yimplies x=tan y,x^2-1=dfrac{sin^2y}{cos^2y}-1=-(1+x^2)cos2y$



                      $$2xcos(2x)+(x^2-1)sin2x=(1+x^2)sin2(arctan x-x)$$



                      $$I=displaystyleintdfrac{2x^2}{2xcos(2x)+(x^2-1)sin(2x)} =intdfrac{2x^2}{(1+x^2)sin2(arctan x-x)}$$



                      Now set $u=arctan x-x,du=-dfrac{x^2 dx}{1+x^2}$



                      $$I=-2intdfrac{du}{sin2u}=?$$



                      Use this







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Feb 2 at 14:33









                      lab bhattacharjeelab bhattacharjee

                      228k15159279




                      228k15159279












                      • $begingroup$
                        $$-2intcsc2u du=-ln(csc2u+cot2u)+K=ln(csc2u-cot2u)+K=ln(tan u)+K$$
                        $endgroup$
                        – lab bhattacharjee
                        Feb 2 at 14:56


















                      • $begingroup$
                        $$-2intcsc2u du=-ln(csc2u+cot2u)+K=ln(csc2u-cot2u)+K=ln(tan u)+K$$
                        $endgroup$
                        – lab bhattacharjee
                        Feb 2 at 14:56
















                      $begingroup$
                      $$-2intcsc2u du=-ln(csc2u+cot2u)+K=ln(csc2u-cot2u)+K=ln(tan u)+K$$
                      $endgroup$
                      – lab bhattacharjee
                      Feb 2 at 14:56




                      $begingroup$
                      $$-2intcsc2u du=-ln(csc2u+cot2u)+K=ln(csc2u-cot2u)+K=ln(tan u)+K$$
                      $endgroup$
                      – lab bhattacharjee
                      Feb 2 at 14:56


















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