Trig substitution problem. Stuck but close.












0












$begingroup$


I have this problem:



$$int_0^2 sqrt{1 + 4x^2}$$



Here's how I start. I focus on finding the antiderivative:
$$ int sqrt{4 * (frac{1}{4} + x^2)}$$



$$ int 2* sqrt{frac{1}{4} + x^2)} dx$$



so let $x = frac{1}{2} * tan{theta}$ so $dx = frac{1}{2} * sec^2theta dtheta$



so...



$$ int 2 sqrt{frac{1}{4} + frac{1}{4}tan^2{theta}} * frac{1}{2}sec^2{theta} cdot dtheta$$



and by $sec^2{theta} = 1 + tan^2{theta}$:



$$ int 2 cdot frac{1}{2} sec{theta} cdot frac{1}{2} sec^2{theta} cdot dtheta$$



$$ int frac{1}{2} sec{theta} cdot sec^2{theta} cdot dtheta$$ Let's call the above line : (1)



and now integration by parts:



$u = sec{theta}$ and so $du = sec{theta}tan{theta}$



$dv = sec^2{theta}$ and so $v = tan{theta}$



so = $$frac{1}{2} (cdot sec{theta}tan{theta} - int tan^2{theta} sec{theta} cdot dtheta)$$



and so since $tan^2{theta} = sec^2{theta} - 1$



$$frac{1}{2} (cdot sec{theta}tan{theta} - int sec^3{theta} - int sec{theta} cdot dtheta)$$



So using (1) we have:



$$(frac{1}{2} cdot sec{theta} cdot tan{theta} - frac{1}{2} int sec^3{theta} - frac{1}{2} int sec{theta} cdot dtheta) = int frac{1}{2} sec{theta} cdot sec^2{theta} cdot dtheta = int frac{1}{2} sec^3{theta} dtheta$$



combining like sec^3{theta} terms:



$$ (frac{1}{2} cdot sec{theta} cdot tan{theta} - frac{1}{2} int sec{theta} cdot dtheta) = sec^3{theta} $$



and now multiplying by $frac{1}{2}$ again:



$$ (frac{1}{4} cdot sec{theta} cdot tan{theta} - frac{1}{4} int sec{theta} cdot dtheta) = frac{1}{2} sec^3{theta} $$



But what's the antiderivative of $int sec{theta} dtheta$ ?



I'm told to do this but I don't really get it:



enter image description here






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    wow. A bit cleaner to use $x = frac{1}{2} sinh t$ although you still need some "double angle" formula.
    $endgroup$
    – Will Jagy
    Jan 30 at 1:17
















0












$begingroup$


I have this problem:



$$int_0^2 sqrt{1 + 4x^2}$$



Here's how I start. I focus on finding the antiderivative:
$$ int sqrt{4 * (frac{1}{4} + x^2)}$$



$$ int 2* sqrt{frac{1}{4} + x^2)} dx$$



so let $x = frac{1}{2} * tan{theta}$ so $dx = frac{1}{2} * sec^2theta dtheta$



so...



$$ int 2 sqrt{frac{1}{4} + frac{1}{4}tan^2{theta}} * frac{1}{2}sec^2{theta} cdot dtheta$$



and by $sec^2{theta} = 1 + tan^2{theta}$:



$$ int 2 cdot frac{1}{2} sec{theta} cdot frac{1}{2} sec^2{theta} cdot dtheta$$



$$ int frac{1}{2} sec{theta} cdot sec^2{theta} cdot dtheta$$ Let's call the above line : (1)



and now integration by parts:



$u = sec{theta}$ and so $du = sec{theta}tan{theta}$



$dv = sec^2{theta}$ and so $v = tan{theta}$



so = $$frac{1}{2} (cdot sec{theta}tan{theta} - int tan^2{theta} sec{theta} cdot dtheta)$$



and so since $tan^2{theta} = sec^2{theta} - 1$



$$frac{1}{2} (cdot sec{theta}tan{theta} - int sec^3{theta} - int sec{theta} cdot dtheta)$$



So using (1) we have:



$$(frac{1}{2} cdot sec{theta} cdot tan{theta} - frac{1}{2} int sec^3{theta} - frac{1}{2} int sec{theta} cdot dtheta) = int frac{1}{2} sec{theta} cdot sec^2{theta} cdot dtheta = int frac{1}{2} sec^3{theta} dtheta$$



combining like sec^3{theta} terms:



$$ (frac{1}{2} cdot sec{theta} cdot tan{theta} - frac{1}{2} int sec{theta} cdot dtheta) = sec^3{theta} $$



and now multiplying by $frac{1}{2}$ again:



$$ (frac{1}{4} cdot sec{theta} cdot tan{theta} - frac{1}{4} int sec{theta} cdot dtheta) = frac{1}{2} sec^3{theta} $$



But what's the antiderivative of $int sec{theta} dtheta$ ?



I'm told to do this but I don't really get it:



enter image description here






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    wow. A bit cleaner to use $x = frac{1}{2} sinh t$ although you still need some "double angle" formula.
    $endgroup$
    – Will Jagy
    Jan 30 at 1:17














0












0








0





$begingroup$


I have this problem:



$$int_0^2 sqrt{1 + 4x^2}$$



Here's how I start. I focus on finding the antiderivative:
$$ int sqrt{4 * (frac{1}{4} + x^2)}$$



$$ int 2* sqrt{frac{1}{4} + x^2)} dx$$



so let $x = frac{1}{2} * tan{theta}$ so $dx = frac{1}{2} * sec^2theta dtheta$



so...



$$ int 2 sqrt{frac{1}{4} + frac{1}{4}tan^2{theta}} * frac{1}{2}sec^2{theta} cdot dtheta$$



and by $sec^2{theta} = 1 + tan^2{theta}$:



$$ int 2 cdot frac{1}{2} sec{theta} cdot frac{1}{2} sec^2{theta} cdot dtheta$$



$$ int frac{1}{2} sec{theta} cdot sec^2{theta} cdot dtheta$$ Let's call the above line : (1)



and now integration by parts:



$u = sec{theta}$ and so $du = sec{theta}tan{theta}$



$dv = sec^2{theta}$ and so $v = tan{theta}$



so = $$frac{1}{2} (cdot sec{theta}tan{theta} - int tan^2{theta} sec{theta} cdot dtheta)$$



and so since $tan^2{theta} = sec^2{theta} - 1$



$$frac{1}{2} (cdot sec{theta}tan{theta} - int sec^3{theta} - int sec{theta} cdot dtheta)$$



So using (1) we have:



$$(frac{1}{2} cdot sec{theta} cdot tan{theta} - frac{1}{2} int sec^3{theta} - frac{1}{2} int sec{theta} cdot dtheta) = int frac{1}{2} sec{theta} cdot sec^2{theta} cdot dtheta = int frac{1}{2} sec^3{theta} dtheta$$



combining like sec^3{theta} terms:



$$ (frac{1}{2} cdot sec{theta} cdot tan{theta} - frac{1}{2} int sec{theta} cdot dtheta) = sec^3{theta} $$



and now multiplying by $frac{1}{2}$ again:



$$ (frac{1}{4} cdot sec{theta} cdot tan{theta} - frac{1}{4} int sec{theta} cdot dtheta) = frac{1}{2} sec^3{theta} $$



But what's the antiderivative of $int sec{theta} dtheta$ ?



I'm told to do this but I don't really get it:



enter image description here






calculus







share|cite|improve this question









$endgroup$




I have this problem:



$$int_0^2 sqrt{1 + 4x^2}$$



Here's how I start. I focus on finding the antiderivative:
$$ int sqrt{4 * (frac{1}{4} + x^2)}$$



$$ int 2* sqrt{frac{1}{4} + x^2)} dx$$



so let $x = frac{1}{2} * tan{theta}$ so $dx = frac{1}{2} * sec^2theta dtheta$



so...



$$ int 2 sqrt{frac{1}{4} + frac{1}{4}tan^2{theta}} * frac{1}{2}sec^2{theta} cdot dtheta$$



and by $sec^2{theta} = 1 + tan^2{theta}$:



$$ int 2 cdot frac{1}{2} sec{theta} cdot frac{1}{2} sec^2{theta} cdot dtheta$$



$$ int frac{1}{2} sec{theta} cdot sec^2{theta} cdot dtheta$$ Let's call the above line : (1)



and now integration by parts:



$u = sec{theta}$ and so $du = sec{theta}tan{theta}$



$dv = sec^2{theta}$ and so $v = tan{theta}$



so = $$frac{1}{2} (cdot sec{theta}tan{theta} - int tan^2{theta} sec{theta} cdot dtheta)$$



and so since $tan^2{theta} = sec^2{theta} - 1$



$$frac{1}{2} (cdot sec{theta}tan{theta} - int sec^3{theta} - int sec{theta} cdot dtheta)$$



So using (1) we have:



$$(frac{1}{2} cdot sec{theta} cdot tan{theta} - frac{1}{2} int sec^3{theta} - frac{1}{2} int sec{theta} cdot dtheta) = int frac{1}{2} sec{theta} cdot sec^2{theta} cdot dtheta = int frac{1}{2} sec^3{theta} dtheta$$



combining like sec^3{theta} terms:



$$ (frac{1}{2} cdot sec{theta} cdot tan{theta} - frac{1}{2} int sec{theta} cdot dtheta) = sec^3{theta} $$



and now multiplying by $frac{1}{2}$ again:



$$ (frac{1}{4} cdot sec{theta} cdot tan{theta} - frac{1}{4} int sec{theta} cdot dtheta) = frac{1}{2} sec^3{theta} $$



But what's the antiderivative of $int sec{theta} dtheta$ ?



I'm told to do this but I don't really get it:



enter image description here






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 30 at 1:13









Jwan622Jwan622

2,34611632




2,34611632












  • $begingroup$
    wow. A bit cleaner to use $x = frac{1}{2} sinh t$ although you still need some "double angle" formula.
    $endgroup$
    – Will Jagy
    Jan 30 at 1:17


















  • $begingroup$
    wow. A bit cleaner to use $x = frac{1}{2} sinh t$ although you still need some "double angle" formula.
    $endgroup$
    – Will Jagy
    Jan 30 at 1:17
















$begingroup$
wow. A bit cleaner to use $x = frac{1}{2} sinh t$ although you still need some "double angle" formula.
$endgroup$
– Will Jagy
Jan 30 at 1:17




$begingroup$
wow. A bit cleaner to use $x = frac{1}{2} sinh t$ although you still need some "double angle" formula.
$endgroup$
– Will Jagy
Jan 30 at 1:17










2 Answers
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$begingroup$

begin{align}
int sec theta , dtheta = int frac{sec^2 theta + sec theta tan theta}{sec theta + tan theta} , dtheta=ln (sec theta + tan theta) + c
end{align}



Note that this is due to the numerator is the derivative of the denominator.



$$int frac{f'(t)}{f(t)} , dt=ln f(t) + c$$






share|cite|improve this answer









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    -1












    $begingroup$

    Using a double-angle formula:



    $$sqrt{17}+frac{1}{4} sinh ^{-1}(4)$$



    enter image description here






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      $begingroup$

      begin{align}
      int sec theta , dtheta = int frac{sec^2 theta + sec theta tan theta}{sec theta + tan theta} , dtheta=ln (sec theta + tan theta) + c
      end{align}



      Note that this is due to the numerator is the derivative of the denominator.



      $$int frac{f'(t)}{f(t)} , dt=ln f(t) + c$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        begin{align}
        int sec theta , dtheta = int frac{sec^2 theta + sec theta tan theta}{sec theta + tan theta} , dtheta=ln (sec theta + tan theta) + c
        end{align}



        Note that this is due to the numerator is the derivative of the denominator.



        $$int frac{f'(t)}{f(t)} , dt=ln f(t) + c$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          begin{align}
          int sec theta , dtheta = int frac{sec^2 theta + sec theta tan theta}{sec theta + tan theta} , dtheta=ln (sec theta + tan theta) + c
          end{align}



          Note that this is due to the numerator is the derivative of the denominator.



          $$int frac{f'(t)}{f(t)} , dt=ln f(t) + c$$






          share|cite|improve this answer









          $endgroup$



          begin{align}
          int sec theta , dtheta = int frac{sec^2 theta + sec theta tan theta}{sec theta + tan theta} , dtheta=ln (sec theta + tan theta) + c
          end{align}



          Note that this is due to the numerator is the derivative of the denominator.



          $$int frac{f'(t)}{f(t)} , dt=ln f(t) + c$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 1:17









          Siong Thye GohSiong Thye Goh

          103k1468120




          103k1468120























              -1












              $begingroup$

              Using a double-angle formula:



              $$sqrt{17}+frac{1}{4} sinh ^{-1}(4)$$



              enter image description here






              share|cite|improve this answer









              $endgroup$


















                -1












                $begingroup$

                Using a double-angle formula:



                $$sqrt{17}+frac{1}{4} sinh ^{-1}(4)$$



                enter image description here






                share|cite|improve this answer









                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  Using a double-angle formula:



                  $$sqrt{17}+frac{1}{4} sinh ^{-1}(4)$$



                  enter image description here






                  share|cite|improve this answer









                  $endgroup$



                  Using a double-angle formula:



                  $$sqrt{17}+frac{1}{4} sinh ^{-1}(4)$$



                  enter image description here







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 30 at 1:18









                  David G. StorkDavid G. Stork

                  11.6k41534




                  11.6k41534






























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