Integrating Inverse Trigs [closed]

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Taking the integral of an inverse trig function by parts.






calculus







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closed as unclear what you're asking by Nosrati, Claude Leibovici, Lord Shark the Unknown, Hans Lundmark, max_zorn Aug 31 at 7:17


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


















    up vote
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    Taking the integral of an inverse trig function by parts.






    calculus







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    closed as unclear what you're asking by Nosrati, Claude Leibovici, Lord Shark the Unknown, Hans Lundmark, max_zorn Aug 31 at 7:17


    Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
















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      Taking the integral of an inverse trig function by parts.






      calculus







      share|cite|improve this question















      Taking the integral of an inverse trig function by parts.






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













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      edited Aug 31 at 5:32

























      asked Aug 31 at 4:58









      Mathman24

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      closed as unclear what you're asking by Nosrati, Claude Leibovici, Lord Shark the Unknown, Hans Lundmark, max_zorn Aug 31 at 7:17


      Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






      closed as unclear what you're asking by Nosrati, Claude Leibovici, Lord Shark the Unknown, Hans Lundmark, max_zorn Aug 31 at 7:17


      Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






















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          Perform instead the substitution $x+1=cosh(t)$.



          In this case we thus have the meaningful primitivation formula



          $$
          int mboxarccosh(x+1)dx=int t~sinh(t)~dt.
          $$



          The conclusion of the exercise can be done afterwards by direct application of integration by parts.






          share|cite|improve this answer






















          • Ahh yes! That seems like a much more fluid method of performing the integration.
            – Mathman24
            Aug 31 at 5:08










          • The method may be comprised on the following identity $$ int f^-1(x)~dx=int t f'(t)~dt$$ underlying to the substitution $t=f(x)$. Here we assume that $f$ is differentiable.
            – Nelson Faustino
            Aug 31 at 5:10











          • Now integrating by parts, there holds that $$ int t f'(t)dt=tf(t)-int f(t)dt.$$
            – Nelson Faustino
            Aug 31 at 5:14

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote













          Perform instead the substitution $x+1=cosh(t)$.



          In this case we thus have the meaningful primitivation formula



          $$
          int mboxarccosh(x+1)dx=int t~sinh(t)~dt.
          $$



          The conclusion of the exercise can be done afterwards by direct application of integration by parts.






          share|cite|improve this answer






















          • Ahh yes! That seems like a much more fluid method of performing the integration.
            – Mathman24
            Aug 31 at 5:08










          • The method may be comprised on the following identity $$ int f^-1(x)~dx=int t f'(t)~dt$$ underlying to the substitution $t=f(x)$. Here we assume that $f$ is differentiable.
            – Nelson Faustino
            Aug 31 at 5:10











          • Now integrating by parts, there holds that $$ int t f'(t)dt=tf(t)-int f(t)dt.$$
            – Nelson Faustino
            Aug 31 at 5:14














          up vote
          1
          down vote













          Perform instead the substitution $x+1=cosh(t)$.



          In this case we thus have the meaningful primitivation formula



          $$
          int mboxarccosh(x+1)dx=int t~sinh(t)~dt.
          $$



          The conclusion of the exercise can be done afterwards by direct application of integration by parts.






          share|cite|improve this answer






















          • Ahh yes! That seems like a much more fluid method of performing the integration.
            – Mathman24
            Aug 31 at 5:08










          • The method may be comprised on the following identity $$ int f^-1(x)~dx=int t f'(t)~dt$$ underlying to the substitution $t=f(x)$. Here we assume that $f$ is differentiable.
            – Nelson Faustino
            Aug 31 at 5:10











          • Now integrating by parts, there holds that $$ int t f'(t)dt=tf(t)-int f(t)dt.$$
            – Nelson Faustino
            Aug 31 at 5:14












          up vote
          1
          down vote










          up vote
          1
          down vote









          Perform instead the substitution $x+1=cosh(t)$.



          In this case we thus have the meaningful primitivation formula



          $$
          int mboxarccosh(x+1)dx=int t~sinh(t)~dt.
          $$



          The conclusion of the exercise can be done afterwards by direct application of integration by parts.






          share|cite|improve this answer














          Perform instead the substitution $x+1=cosh(t)$.



          In this case we thus have the meaningful primitivation formula



          $$
          int mboxarccosh(x+1)dx=int t~sinh(t)~dt.
          $$



          The conclusion of the exercise can be done afterwards by direct application of integration by parts.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 31 at 5:16

























          answered Aug 31 at 5:06









          Nelson Faustino

          1376




          1376











          • Ahh yes! That seems like a much more fluid method of performing the integration.
            – Mathman24
            Aug 31 at 5:08










          • The method may be comprised on the following identity $$ int f^-1(x)~dx=int t f'(t)~dt$$ underlying to the substitution $t=f(x)$. Here we assume that $f$ is differentiable.
            – Nelson Faustino
            Aug 31 at 5:10











          • Now integrating by parts, there holds that $$ int t f'(t)dt=tf(t)-int f(t)dt.$$
            – Nelson Faustino
            Aug 31 at 5:14
















          • Ahh yes! That seems like a much more fluid method of performing the integration.
            – Mathman24
            Aug 31 at 5:08










          • The method may be comprised on the following identity $$ int f^-1(x)~dx=int t f'(t)~dt$$ underlying to the substitution $t=f(x)$. Here we assume that $f$ is differentiable.
            – Nelson Faustino
            Aug 31 at 5:10











          • Now integrating by parts, there holds that $$ int t f'(t)dt=tf(t)-int f(t)dt.$$
            – Nelson Faustino
            Aug 31 at 5:14















          Ahh yes! That seems like a much more fluid method of performing the integration.
          – Mathman24
          Aug 31 at 5:08




          Ahh yes! That seems like a much more fluid method of performing the integration.
          – Mathman24
          Aug 31 at 5:08












          The method may be comprised on the following identity $$ int f^-1(x)~dx=int t f'(t)~dt$$ underlying to the substitution $t=f(x)$. Here we assume that $f$ is differentiable.
          – Nelson Faustino
          Aug 31 at 5:10





          The method may be comprised on the following identity $$ int f^-1(x)~dx=int t f'(t)~dt$$ underlying to the substitution $t=f(x)$. Here we assume that $f$ is differentiable.
          – Nelson Faustino
          Aug 31 at 5:10













          Now integrating by parts, there holds that $$ int t f'(t)dt=tf(t)-int f(t)dt.$$
          – Nelson Faustino
          Aug 31 at 5:14




          Now integrating by parts, there holds that $$ int t f'(t)dt=tf(t)-int f(t)dt.$$
          – Nelson Faustino
          Aug 31 at 5:14


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