Integrate using trigonometric substitution. Am I on the right path?












3












$begingroup$


I have been trying to solve:



$$int frac{sqrt{x^2-9}}{x^3} dx$$



I am letting $ x = 3sec theta$ and so $dx = 3 sec theta tan theta$



So then I have:



$$int frac{sqrt{9sec^2 theta - 9}}{27 sec^3 theta} dx$$



$$int frac{sqrt{9(sec^2 theta - 1)}}{27 sec^3 theta} 3 sec theta tan theta, d theta$$



$$ int frac{3tan theta}{27 sec^3 theta} 3 sec theta tan theta, d theta$$



$$ = int frac{9 tan ^2 theta}{27 sec ^2 theta} d theta$$



$$ = int frac{tan ^2 theta}{3 sec ^2 theta} d theta$$



$$ int frac{sin^2 theta}{cos^2 theta} cdot frac{cos^2 theta}{3} d theta$$



$$ int frac{sin^2 theta}{3} d theta$$



$$frac{1}{3} int sin^2 theta d theta$$



Am I on the right track?






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
    $endgroup$
    – Bernard
    Feb 3 at 16:46








  • 1




    $begingroup$
    On the right track except for some missing $=$ and $dtheta$.
    $endgroup$
    – Bernard Massé
    Feb 3 at 17:06


















3












$begingroup$


I have been trying to solve:



$$int frac{sqrt{x^2-9}}{x^3} dx$$



I am letting $ x = 3sec theta$ and so $dx = 3 sec theta tan theta$



So then I have:



$$int frac{sqrt{9sec^2 theta - 9}}{27 sec^3 theta} dx$$



$$int frac{sqrt{9(sec^2 theta - 1)}}{27 sec^3 theta} 3 sec theta tan theta, d theta$$



$$ int frac{3tan theta}{27 sec^3 theta} 3 sec theta tan theta, d theta$$



$$ = int frac{9 tan ^2 theta}{27 sec ^2 theta} d theta$$



$$ = int frac{tan ^2 theta}{3 sec ^2 theta} d theta$$



$$ int frac{sin^2 theta}{cos^2 theta} cdot frac{cos^2 theta}{3} d theta$$



$$ int frac{sin^2 theta}{3} d theta$$



$$frac{1}{3} int sin^2 theta d theta$$



Am I on the right track?






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
    $endgroup$
    – Bernard
    Feb 3 at 16:46








  • 1




    $begingroup$
    On the right track except for some missing $=$ and $dtheta$.
    $endgroup$
    – Bernard Massé
    Feb 3 at 17:06
















3












3








3





$begingroup$


I have been trying to solve:



$$int frac{sqrt{x^2-9}}{x^3} dx$$



I am letting $ x = 3sec theta$ and so $dx = 3 sec theta tan theta$



So then I have:



$$int frac{sqrt{9sec^2 theta - 9}}{27 sec^3 theta} dx$$



$$int frac{sqrt{9(sec^2 theta - 1)}}{27 sec^3 theta} 3 sec theta tan theta, d theta$$



$$ int frac{3tan theta}{27 sec^3 theta} 3 sec theta tan theta, d theta$$



$$ = int frac{9 tan ^2 theta}{27 sec ^2 theta} d theta$$



$$ = int frac{tan ^2 theta}{3 sec ^2 theta} d theta$$



$$ int frac{sin^2 theta}{cos^2 theta} cdot frac{cos^2 theta}{3} d theta$$



$$ int frac{sin^2 theta}{3} d theta$$



$$frac{1}{3} int sin^2 theta d theta$$



Am I on the right track?






integration







share|cite|improve this question











$endgroup$




I have been trying to solve:



$$int frac{sqrt{x^2-9}}{x^3} dx$$



I am letting $ x = 3sec theta$ and so $dx = 3 sec theta tan theta$



So then I have:



$$int frac{sqrt{9sec^2 theta - 9}}{27 sec^3 theta} dx$$



$$int frac{sqrt{9(sec^2 theta - 1)}}{27 sec^3 theta} 3 sec theta tan theta, d theta$$



$$ int frac{3tan theta}{27 sec^3 theta} 3 sec theta tan theta, d theta$$



$$ = int frac{9 tan ^2 theta}{27 sec ^2 theta} d theta$$



$$ = int frac{tan ^2 theta}{3 sec ^2 theta} d theta$$



$$ int frac{sin^2 theta}{cos^2 theta} cdot frac{cos^2 theta}{3} d theta$$



$$ int frac{sin^2 theta}{3} d theta$$



$$frac{1}{3} int sin^2 theta d theta$$



Am I on the right track?






integration




integration






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share|cite|improve this question













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edited Feb 3 at 17:17









J. W. Tanner

3,6551320




3,6551320










asked Feb 3 at 16:36









Jwan622Jwan622

2,28011632




2,28011632












  • $begingroup$
    Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
    $endgroup$
    – Bernard
    Feb 3 at 16:46








  • 1




    $begingroup$
    On the right track except for some missing $=$ and $dtheta$.
    $endgroup$
    – Bernard Massé
    Feb 3 at 17:06




















  • $begingroup$
    Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
    $endgroup$
    – Bernard
    Feb 3 at 16:46








  • 1




    $begingroup$
    On the right track except for some missing $=$ and $dtheta$.
    $endgroup$
    – Bernard Massé
    Feb 3 at 17:06


















$begingroup$
Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
$endgroup$
– Bernard
Feb 3 at 16:46






$begingroup$
Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
$endgroup$
– Bernard
Feb 3 at 16:46






1




1




$begingroup$
On the right track except for some missing $=$ and $dtheta$.
$endgroup$
– Bernard Massé
Feb 3 at 17:06






$begingroup$
On the right track except for some missing $=$ and $dtheta$.
$endgroup$
– Bernard Massé
Feb 3 at 17:06












2 Answers
2






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4












$begingroup$

Yes, your solution so far is correct. Now, to integrate $int sin^2{theta},dtheta$, use the half-angle formula for the sine function:



$$
sin^2{theta}=frac{1-cos(2theta)}{2}.
$$



Also, a bit later, you're going to need this formula:



$$
sin{(2theta)}=2sin{theta}cos{theta}.
$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $$I=intfrac{sqrt{x^2-9}}{x^3}dx=intfrac{3sqrt{(x/3)^2-1}}{x^3}dx$$
    and we know that $cosh^2theta-1=sinh^2theta$
    $$x=3cosh(y)$$
    $$dx=3sinh(y)dy$$
    $$I=9intfrac{sqrt{cosh^2(y)-1}}{27cosh^3(y)}sinh(y)dy=frac13intfrac{sinh^2(y)}{cosh^3(y)}dy=frac13inttext{sech}(y)-text{sech}^3(y)dy$$
    a reduction formula can then be used






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      4












      $begingroup$

      Yes, your solution so far is correct. Now, to integrate $int sin^2{theta},dtheta$, use the half-angle formula for the sine function:



      $$
      sin^2{theta}=frac{1-cos(2theta)}{2}.
      $$



      Also, a bit later, you're going to need this formula:



      $$
      sin{(2theta)}=2sin{theta}cos{theta}.
      $$






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        Yes, your solution so far is correct. Now, to integrate $int sin^2{theta},dtheta$, use the half-angle formula for the sine function:



        $$
        sin^2{theta}=frac{1-cos(2theta)}{2}.
        $$



        Also, a bit later, you're going to need this formula:



        $$
        sin{(2theta)}=2sin{theta}cos{theta}.
        $$






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          Yes, your solution so far is correct. Now, to integrate $int sin^2{theta},dtheta$, use the half-angle formula for the sine function:



          $$
          sin^2{theta}=frac{1-cos(2theta)}{2}.
          $$



          Also, a bit later, you're going to need this formula:



          $$
          sin{(2theta)}=2sin{theta}cos{theta}.
          $$






          share|cite|improve this answer











          $endgroup$



          Yes, your solution so far is correct. Now, to integrate $int sin^2{theta},dtheta$, use the half-angle formula for the sine function:



          $$
          sin^2{theta}=frac{1-cos(2theta)}{2}.
          $$



          Also, a bit later, you're going to need this formula:



          $$
          sin{(2theta)}=2sin{theta}cos{theta}.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 3 at 17:15

























          answered Feb 3 at 17:02









          Michael RybkinMichael Rybkin

          3,939421




          3,939421























              1












              $begingroup$

              $$I=intfrac{sqrt{x^2-9}}{x^3}dx=intfrac{3sqrt{(x/3)^2-1}}{x^3}dx$$
              and we know that $cosh^2theta-1=sinh^2theta$
              $$x=3cosh(y)$$
              $$dx=3sinh(y)dy$$
              $$I=9intfrac{sqrt{cosh^2(y)-1}}{27cosh^3(y)}sinh(y)dy=frac13intfrac{sinh^2(y)}{cosh^3(y)}dy=frac13inttext{sech}(y)-text{sech}^3(y)dy$$
              a reduction formula can then be used






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $$I=intfrac{sqrt{x^2-9}}{x^3}dx=intfrac{3sqrt{(x/3)^2-1}}{x^3}dx$$
                and we know that $cosh^2theta-1=sinh^2theta$
                $$x=3cosh(y)$$
                $$dx=3sinh(y)dy$$
                $$I=9intfrac{sqrt{cosh^2(y)-1}}{27cosh^3(y)}sinh(y)dy=frac13intfrac{sinh^2(y)}{cosh^3(y)}dy=frac13inttext{sech}(y)-text{sech}^3(y)dy$$
                a reduction formula can then be used






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $$I=intfrac{sqrt{x^2-9}}{x^3}dx=intfrac{3sqrt{(x/3)^2-1}}{x^3}dx$$
                  and we know that $cosh^2theta-1=sinh^2theta$
                  $$x=3cosh(y)$$
                  $$dx=3sinh(y)dy$$
                  $$I=9intfrac{sqrt{cosh^2(y)-1}}{27cosh^3(y)}sinh(y)dy=frac13intfrac{sinh^2(y)}{cosh^3(y)}dy=frac13inttext{sech}(y)-text{sech}^3(y)dy$$
                  a reduction formula can then be used






                  share|cite|improve this answer









                  $endgroup$



                  $$I=intfrac{sqrt{x^2-9}}{x^3}dx=intfrac{3sqrt{(x/3)^2-1}}{x^3}dx$$
                  and we know that $cosh^2theta-1=sinh^2theta$
                  $$x=3cosh(y)$$
                  $$dx=3sinh(y)dy$$
                  $$I=9intfrac{sqrt{cosh^2(y)-1}}{27cosh^3(y)}sinh(y)dy=frac13intfrac{sinh^2(y)}{cosh^3(y)}dy=frac13inttext{sech}(y)-text{sech}^3(y)dy$$
                  a reduction formula can then be used







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 3 at 20:29









                  Henry LeeHenry Lee

                  2,199319




                  2,199319






























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