Applying substitution to $int sqrt{1+sin(x)}$ [duplicate]












0












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This question already has an answer here:




  • How to evaluate the integral $int sqrt{1+sin(x)} dx$

    6 answers




I'm having a problem with substituting when it comes to trig functions.



The integral is:$$intsqrt{1+sin(x)}dx$$



Substituting: $1+sin(x) = u implies frac{d}{dx} u = cos(x) implies dx = frac{1}{cos(x)}du$



So now we have the integral: $$int frac{sqrt{u}}{cos(x)}du$$



So now the question is, what do I do with the $frac{1}{cos(x)}$? How can I substitute that with $u$?






integration substitution







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marked as duplicate by clathratus, stressed out, Lord Shark the Unknown, Abcd, The Chaz 2.0 Jan 4 at 6:34


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    Note $cos(x) = sqrt{1 - sin^2(x)} = sqrt{1 - (u-1)^2} $ as the general method. But maybe you find a nicer (similar) substitution ...
    $endgroup$
    – Andreas
    Jan 3 at 21:53












  • $begingroup$
    Hint: $frac{1+cos(2x)}2=cos^2(x)$ and $sin(x)=cosleft(fracpi2-xright)$
    $endgroup$
    – robjohn♦
    Jan 3 at 21:53








  • 1




    $begingroup$
    I wouldn't go for this substitution, but use the identity$$sin^2theta+cos^2theta=1$$to get back in terms of $u$
    $endgroup$
    – Frank W.
    Jan 3 at 21:53










  • $begingroup$
    It is $$cos(x)=pm sqrt{1-sin^2(x)}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 21:56


















0












$begingroup$



This question already has an answer here:




  • How to evaluate the integral $int sqrt{1+sin(x)} dx$

    6 answers




I'm having a problem with substituting when it comes to trig functions.



The integral is:$$intsqrt{1+sin(x)}dx$$



Substituting: $1+sin(x) = u implies frac{d}{dx} u = cos(x) implies dx = frac{1}{cos(x)}du$



So now we have the integral: $$int frac{sqrt{u}}{cos(x)}du$$



So now the question is, what do I do with the $frac{1}{cos(x)}$? How can I substitute that with $u$?






integration substitution







share|cite|improve this question











$endgroup$



marked as duplicate by clathratus, stressed out, Lord Shark the Unknown, Abcd, The Chaz 2.0 Jan 4 at 6:34


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    Note $cos(x) = sqrt{1 - sin^2(x)} = sqrt{1 - (u-1)^2} $ as the general method. But maybe you find a nicer (similar) substitution ...
    $endgroup$
    – Andreas
    Jan 3 at 21:53












  • $begingroup$
    Hint: $frac{1+cos(2x)}2=cos^2(x)$ and $sin(x)=cosleft(fracpi2-xright)$
    $endgroup$
    – robjohn♦
    Jan 3 at 21:53








  • 1




    $begingroup$
    I wouldn't go for this substitution, but use the identity$$sin^2theta+cos^2theta=1$$to get back in terms of $u$
    $endgroup$
    – Frank W.
    Jan 3 at 21:53










  • $begingroup$
    It is $$cos(x)=pm sqrt{1-sin^2(x)}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 21:56
















0












0








0





$begingroup$



This question already has an answer here:




  • How to evaluate the integral $int sqrt{1+sin(x)} dx$

    6 answers




I'm having a problem with substituting when it comes to trig functions.



The integral is:$$intsqrt{1+sin(x)}dx$$



Substituting: $1+sin(x) = u implies frac{d}{dx} u = cos(x) implies dx = frac{1}{cos(x)}du$



So now we have the integral: $$int frac{sqrt{u}}{cos(x)}du$$



So now the question is, what do I do with the $frac{1}{cos(x)}$? How can I substitute that with $u$?






integration substitution







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • How to evaluate the integral $int sqrt{1+sin(x)} dx$

    6 answers




I'm having a problem with substituting when it comes to trig functions.



The integral is:$$intsqrt{1+sin(x)}dx$$



Substituting: $1+sin(x) = u implies frac{d}{dx} u = cos(x) implies dx = frac{1}{cos(x)}du$



So now we have the integral: $$int frac{sqrt{u}}{cos(x)}du$$



So now the question is, what do I do with the $frac{1}{cos(x)}$? How can I substitute that with $u$?





This question already has an answer here:




  • How to evaluate the integral $int sqrt{1+sin(x)} dx$

    6 answers







integration substitution




integration substitution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 22:01









gt6989b

33.7k22455




33.7k22455










asked Jan 3 at 21:49









Conny DagoConny Dago

225




225




marked as duplicate by clathratus, stressed out, Lord Shark the Unknown, Abcd, The Chaz 2.0 Jan 4 at 6:34


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by clathratus, stressed out, Lord Shark the Unknown, Abcd, The Chaz 2.0 Jan 4 at 6:34


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Note $cos(x) = sqrt{1 - sin^2(x)} = sqrt{1 - (u-1)^2} $ as the general method. But maybe you find a nicer (similar) substitution ...
    $endgroup$
    – Andreas
    Jan 3 at 21:53












  • $begingroup$
    Hint: $frac{1+cos(2x)}2=cos^2(x)$ and $sin(x)=cosleft(fracpi2-xright)$
    $endgroup$
    – robjohn♦
    Jan 3 at 21:53








  • 1




    $begingroup$
    I wouldn't go for this substitution, but use the identity$$sin^2theta+cos^2theta=1$$to get back in terms of $u$
    $endgroup$
    – Frank W.
    Jan 3 at 21:53










  • $begingroup$
    It is $$cos(x)=pm sqrt{1-sin^2(x)}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 21:56




















  • $begingroup$
    Note $cos(x) = sqrt{1 - sin^2(x)} = sqrt{1 - (u-1)^2} $ as the general method. But maybe you find a nicer (similar) substitution ...
    $endgroup$
    – Andreas
    Jan 3 at 21:53












  • $begingroup$
    Hint: $frac{1+cos(2x)}2=cos^2(x)$ and $sin(x)=cosleft(fracpi2-xright)$
    $endgroup$
    – robjohn♦
    Jan 3 at 21:53








  • 1




    $begingroup$
    I wouldn't go for this substitution, but use the identity$$sin^2theta+cos^2theta=1$$to get back in terms of $u$
    $endgroup$
    – Frank W.
    Jan 3 at 21:53










  • $begingroup$
    It is $$cos(x)=pm sqrt{1-sin^2(x)}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 21:56


















$begingroup$
Note $cos(x) = sqrt{1 - sin^2(x)} = sqrt{1 - (u-1)^2} $ as the general method. But maybe you find a nicer (similar) substitution ...
$endgroup$
– Andreas
Jan 3 at 21:53






$begingroup$
Note $cos(x) = sqrt{1 - sin^2(x)} = sqrt{1 - (u-1)^2} $ as the general method. But maybe you find a nicer (similar) substitution ...
$endgroup$
– Andreas
Jan 3 at 21:53














$begingroup$
Hint: $frac{1+cos(2x)}2=cos^2(x)$ and $sin(x)=cosleft(fracpi2-xright)$
$endgroup$
– robjohn♦
Jan 3 at 21:53






$begingroup$
Hint: $frac{1+cos(2x)}2=cos^2(x)$ and $sin(x)=cosleft(fracpi2-xright)$
$endgroup$
– robjohn♦
Jan 3 at 21:53






1




1




$begingroup$
I wouldn't go for this substitution, but use the identity$$sin^2theta+cos^2theta=1$$to get back in terms of $u$
$endgroup$
– Frank W.
Jan 3 at 21:53




$begingroup$
I wouldn't go for this substitution, but use the identity$$sin^2theta+cos^2theta=1$$to get back in terms of $u$
$endgroup$
– Frank W.
Jan 3 at 21:53












$begingroup$
It is $$cos(x)=pm sqrt{1-sin^2(x)}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 21:56






$begingroup$
It is $$cos(x)=pm sqrt{1-sin^2(x)}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 21:56












3 Answers
3






active

oldest

votes


















3












$begingroup$

Hint Write your integrand in the form
$$sqrt{frac{(1+sin(x))(1-sin(x))}{1- sin(x)}}=frac{pmcos(x)}{sqrt{1-sin(x)}}$$
and substitute $$t=1-sin(x)$$ then we get $$dt=-cos(x)dx$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    HINT



    Here I give you an alternative approach. Observer that



    begin{align*}
    1 + sin(x) & = 1 + 2sinleft(frac{x}{2}right)cosleft(frac{x}{2}right) = left[cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right]^{2}
    end{align*}



    Hence the given integral is equal to



    begin{align*}
    intsqrt{1+sin(x)}mathrm{d}x = intleft|cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right|mathrm{d}x
    end{align*}






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint:
      $$
      begin{align}
      sqrt{frac{1+sin(x)}2}
      &=sqrt{frac{1+cosleft(fracpi2-xright)}2}\
      &=left|,cosleft(tfracpi4-tfrac x2right),right|
      end{align}
      $$






      share|cite|improve this answer









      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Hint Write your integrand in the form
        $$sqrt{frac{(1+sin(x))(1-sin(x))}{1- sin(x)}}=frac{pmcos(x)}{sqrt{1-sin(x)}}$$
        and substitute $$t=1-sin(x)$$ then we get $$dt=-cos(x)dx$$






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Hint Write your integrand in the form
          $$sqrt{frac{(1+sin(x))(1-sin(x))}{1- sin(x)}}=frac{pmcos(x)}{sqrt{1-sin(x)}}$$
          and substitute $$t=1-sin(x)$$ then we get $$dt=-cos(x)dx$$






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Hint Write your integrand in the form
            $$sqrt{frac{(1+sin(x))(1-sin(x))}{1- sin(x)}}=frac{pmcos(x)}{sqrt{1-sin(x)}}$$
            and substitute $$t=1-sin(x)$$ then we get $$dt=-cos(x)dx$$






            share|cite|improve this answer









            $endgroup$



            Hint Write your integrand in the form
            $$sqrt{frac{(1+sin(x))(1-sin(x))}{1- sin(x)}}=frac{pmcos(x)}{sqrt{1-sin(x)}}$$
            and substitute $$t=1-sin(x)$$ then we get $$dt=-cos(x)dx$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 3 at 21:56









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            73.7k42865




            73.7k42865























                2












                $begingroup$

                HINT



                Here I give you an alternative approach. Observer that



                begin{align*}
                1 + sin(x) & = 1 + 2sinleft(frac{x}{2}right)cosleft(frac{x}{2}right) = left[cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right]^{2}
                end{align*}



                Hence the given integral is equal to



                begin{align*}
                intsqrt{1+sin(x)}mathrm{d}x = intleft|cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right|mathrm{d}x
                end{align*}






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  HINT



                  Here I give you an alternative approach. Observer that



                  begin{align*}
                  1 + sin(x) & = 1 + 2sinleft(frac{x}{2}right)cosleft(frac{x}{2}right) = left[cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right]^{2}
                  end{align*}



                  Hence the given integral is equal to



                  begin{align*}
                  intsqrt{1+sin(x)}mathrm{d}x = intleft|cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right|mathrm{d}x
                  end{align*}






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    HINT



                    Here I give you an alternative approach. Observer that



                    begin{align*}
                    1 + sin(x) & = 1 + 2sinleft(frac{x}{2}right)cosleft(frac{x}{2}right) = left[cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right]^{2}
                    end{align*}



                    Hence the given integral is equal to



                    begin{align*}
                    intsqrt{1+sin(x)}mathrm{d}x = intleft|cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right|mathrm{d}x
                    end{align*}






                    share|cite|improve this answer









                    $endgroup$



                    HINT



                    Here I give you an alternative approach. Observer that



                    begin{align*}
                    1 + sin(x) & = 1 + 2sinleft(frac{x}{2}right)cosleft(frac{x}{2}right) = left[cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right]^{2}
                    end{align*}



                    Hence the given integral is equal to



                    begin{align*}
                    intsqrt{1+sin(x)}mathrm{d}x = intleft|cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right|mathrm{d}x
                    end{align*}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 3 at 21:59









                    APC89APC89

                    2,016418




                    2,016418























                        1












                        $begingroup$

                        Hint:
                        $$
                        begin{align}
                        sqrt{frac{1+sin(x)}2}
                        &=sqrt{frac{1+cosleft(fracpi2-xright)}2}\
                        &=left|,cosleft(tfracpi4-tfrac x2right),right|
                        end{align}
                        $$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint:
                          $$
                          begin{align}
                          sqrt{frac{1+sin(x)}2}
                          &=sqrt{frac{1+cosleft(fracpi2-xright)}2}\
                          &=left|,cosleft(tfracpi4-tfrac x2right),right|
                          end{align}
                          $$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint:
                            $$
                            begin{align}
                            sqrt{frac{1+sin(x)}2}
                            &=sqrt{frac{1+cosleft(fracpi2-xright)}2}\
                            &=left|,cosleft(tfracpi4-tfrac x2right),right|
                            end{align}
                            $$






                            share|cite|improve this answer









                            $endgroup$



                            Hint:
                            $$
                            begin{align}
                            sqrt{frac{1+sin(x)}2}
                            &=sqrt{frac{1+cosleft(fracpi2-xright)}2}\
                            &=left|,cosleft(tfracpi4-tfrac x2right),right|
                            end{align}
                            $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 3 at 22:02









                            robjohn♦robjohn

                            266k27304626




                            266k27304626















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