Prove that...

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One of the problems in my homework set ask me to prove the following identity:
$$int{sec^{n}(theta)}dtheta=frac{tan(theta)sec^{n-2}(theta)}{n-1}-frac{n-2}{n-1}int{sec^{n-2}(theta)dtheta}$$
Can someone give me a hint how to start?
Thanks!






calculus integration reduction-formula







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  • $begingroup$
    Smelly like integration by parts.
    $endgroup$
    – Rasmus
    Aug 9 '13 at 14:42
















0












$begingroup$


One of the problems in my homework set ask me to prove the following identity:
$$int{sec^{n}(theta)}dtheta=frac{tan(theta)sec^{n-2}(theta)}{n-1}-frac{n-2}{n-1}int{sec^{n-2}(theta)dtheta}$$
Can someone give me a hint how to start?
Thanks!






calculus integration reduction-formula







share|cite|improve this question











$endgroup$












  • $begingroup$
    Smelly like integration by parts.
    $endgroup$
    – Rasmus
    Aug 9 '13 at 14:42














0












0








0





$begingroup$


One of the problems in my homework set ask me to prove the following identity:
$$int{sec^{n}(theta)}dtheta=frac{tan(theta)sec^{n-2}(theta)}{n-1}-frac{n-2}{n-1}int{sec^{n-2}(theta)dtheta}$$
Can someone give me a hint how to start?
Thanks!






calculus integration reduction-formula







share|cite|improve this question











$endgroup$




One of the problems in my homework set ask me to prove the following identity:
$$int{sec^{n}(theta)}dtheta=frac{tan(theta)sec^{n-2}(theta)}{n-1}-frac{n-2}{n-1}int{sec^{n-2}(theta)dtheta}$$
Can someone give me a hint how to start?
Thanks!






calculus integration reduction-formula




calculus integration reduction-formula






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edited Jan 6 at 23:08









clathratus

3,484331




3,484331










asked Aug 9 '13 at 14:37







user89632



















  • $begingroup$
    Smelly like integration by parts.
    $endgroup$
    – Rasmus
    Aug 9 '13 at 14:42


















  • $begingroup$
    Smelly like integration by parts.
    $endgroup$
    – Rasmus
    Aug 9 '13 at 14:42
















$begingroup$
Smelly like integration by parts.
$endgroup$
– Rasmus
Aug 9 '13 at 14:42




$begingroup$
Smelly like integration by parts.
$endgroup$
– Rasmus
Aug 9 '13 at 14:42










4 Answers
4






active

oldest

votes


















3












$begingroup$

Using integration by parts,
$$intsec^nxdx=intsec^{n-2}xcdot sec^2xdx$$
$$=sec^{n-2}xintsec^2xdx-intleft(frac{d(sec^{n-2}x)}{dx}cdot sec^2xdxright)dx $$



$$=sec^{n-2}xtan x-intleft((n-2)sec^{n-3}x(sec xtan x)cdot tan xright)dx $$



$$=sec^{n-2}xtan x-(n-2)intsec^{n-2}x(sec^2x-1)dx $$



$$=sec^{n-2}xtan x-(n-2)intsec^nxdx+(n-2)intsec^{n-2}xdx+C $$ where $C$ is an arbitrary constant for indefinite integration



$$implies (1+n-2)sec^nxdx=sec^{n-2}xtan x+(n-2)intsec^{n-2}xdx+C$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How about the +C?
    $endgroup$
    – user89632
    Aug 9 '13 at 14:47










  • $begingroup$
    @user89632, added the arbitrary constant, but the please check the problem. It's $+(n-2)sec^{n-2}xdx$ also replace $theta$ with $x$
    $endgroup$
    – lab bhattacharjee
    Aug 9 '13 at 14:53



















0












$begingroup$

Hint:



Use the fact that



$$tan^2 (theta)+1=sec^2(theta)$$



Now , $sec^n theta=sec^{n-2} theta cdot sec^2 theta=sec^{n-2}theta cdot (tan^2 theta+1)$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint:



    $$int{sec^{n}(theta)}dtheta= int{sec^{n-2}(theta)} sec^2(theta) dtheta$$



    Integrate by parts.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Hint:



      remember that $d(tan theta) = frac{dtheta}{cos^2theta} = sec^2 theta dtheta$, integrate by parts, then collect the like terms with $int frac{dtheta}{cos^n theta}$






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Using integration by parts,
        $$intsec^nxdx=intsec^{n-2}xcdot sec^2xdx$$
        $$=sec^{n-2}xintsec^2xdx-intleft(frac{d(sec^{n-2}x)}{dx}cdot sec^2xdxright)dx $$



        $$=sec^{n-2}xtan x-intleft((n-2)sec^{n-3}x(sec xtan x)cdot tan xright)dx $$



        $$=sec^{n-2}xtan x-(n-2)intsec^{n-2}x(sec^2x-1)dx $$



        $$=sec^{n-2}xtan x-(n-2)intsec^nxdx+(n-2)intsec^{n-2}xdx+C $$ where $C$ is an arbitrary constant for indefinite integration



        $$implies (1+n-2)sec^nxdx=sec^{n-2}xtan x+(n-2)intsec^{n-2}xdx+C$$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          How about the +C?
          $endgroup$
          – user89632
          Aug 9 '13 at 14:47










        • $begingroup$
          @user89632, added the arbitrary constant, but the please check the problem. It's $+(n-2)sec^{n-2}xdx$ also replace $theta$ with $x$
          $endgroup$
          – lab bhattacharjee
          Aug 9 '13 at 14:53
















        3












        $begingroup$

        Using integration by parts,
        $$intsec^nxdx=intsec^{n-2}xcdot sec^2xdx$$
        $$=sec^{n-2}xintsec^2xdx-intleft(frac{d(sec^{n-2}x)}{dx}cdot sec^2xdxright)dx $$



        $$=sec^{n-2}xtan x-intleft((n-2)sec^{n-3}x(sec xtan x)cdot tan xright)dx $$



        $$=sec^{n-2}xtan x-(n-2)intsec^{n-2}x(sec^2x-1)dx $$



        $$=sec^{n-2}xtan x-(n-2)intsec^nxdx+(n-2)intsec^{n-2}xdx+C $$ where $C$ is an arbitrary constant for indefinite integration



        $$implies (1+n-2)sec^nxdx=sec^{n-2}xtan x+(n-2)intsec^{n-2}xdx+C$$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          How about the +C?
          $endgroup$
          – user89632
          Aug 9 '13 at 14:47










        • $begingroup$
          @user89632, added the arbitrary constant, but the please check the problem. It's $+(n-2)sec^{n-2}xdx$ also replace $theta$ with $x$
          $endgroup$
          – lab bhattacharjee
          Aug 9 '13 at 14:53














        3












        3








        3





        $begingroup$

        Using integration by parts,
        $$intsec^nxdx=intsec^{n-2}xcdot sec^2xdx$$
        $$=sec^{n-2}xintsec^2xdx-intleft(frac{d(sec^{n-2}x)}{dx}cdot sec^2xdxright)dx $$



        $$=sec^{n-2}xtan x-intleft((n-2)sec^{n-3}x(sec xtan x)cdot tan xright)dx $$



        $$=sec^{n-2}xtan x-(n-2)intsec^{n-2}x(sec^2x-1)dx $$



        $$=sec^{n-2}xtan x-(n-2)intsec^nxdx+(n-2)intsec^{n-2}xdx+C $$ where $C$ is an arbitrary constant for indefinite integration



        $$implies (1+n-2)sec^nxdx=sec^{n-2}xtan x+(n-2)intsec^{n-2}xdx+C$$






        share|cite|improve this answer











        $endgroup$



        Using integration by parts,
        $$intsec^nxdx=intsec^{n-2}xcdot sec^2xdx$$
        $$=sec^{n-2}xintsec^2xdx-intleft(frac{d(sec^{n-2}x)}{dx}cdot sec^2xdxright)dx $$



        $$=sec^{n-2}xtan x-intleft((n-2)sec^{n-3}x(sec xtan x)cdot tan xright)dx $$



        $$=sec^{n-2}xtan x-(n-2)intsec^{n-2}x(sec^2x-1)dx $$



        $$=sec^{n-2}xtan x-(n-2)intsec^nxdx+(n-2)intsec^{n-2}xdx+C $$ where $C$ is an arbitrary constant for indefinite integration



        $$implies (1+n-2)sec^nxdx=sec^{n-2}xtan x+(n-2)intsec^{n-2}xdx+C$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 9 '13 at 14:51

























        answered Aug 9 '13 at 14:42









        lab bhattacharjeelab bhattacharjee

        224k15156274




        224k15156274












        • $begingroup$
          How about the +C?
          $endgroup$
          – user89632
          Aug 9 '13 at 14:47










        • $begingroup$
          @user89632, added the arbitrary constant, but the please check the problem. It's $+(n-2)sec^{n-2}xdx$ also replace $theta$ with $x$
          $endgroup$
          – lab bhattacharjee
          Aug 9 '13 at 14:53


















        • $begingroup$
          How about the +C?
          $endgroup$
          – user89632
          Aug 9 '13 at 14:47










        • $begingroup$
          @user89632, added the arbitrary constant, but the please check the problem. It's $+(n-2)sec^{n-2}xdx$ also replace $theta$ with $x$
          $endgroup$
          – lab bhattacharjee
          Aug 9 '13 at 14:53
















        $begingroup$
        How about the +C?
        $endgroup$
        – user89632
        Aug 9 '13 at 14:47




        $begingroup$
        How about the +C?
        $endgroup$
        – user89632
        Aug 9 '13 at 14:47












        $begingroup$
        @user89632, added the arbitrary constant, but the please check the problem. It's $+(n-2)sec^{n-2}xdx$ also replace $theta$ with $x$
        $endgroup$
        – lab bhattacharjee
        Aug 9 '13 at 14:53




        $begingroup$
        @user89632, added the arbitrary constant, but the please check the problem. It's $+(n-2)sec^{n-2}xdx$ also replace $theta$ with $x$
        $endgroup$
        – lab bhattacharjee
        Aug 9 '13 at 14:53











        0












        $begingroup$

        Hint:



        Use the fact that



        $$tan^2 (theta)+1=sec^2(theta)$$



        Now , $sec^n theta=sec^{n-2} theta cdot sec^2 theta=sec^{n-2}theta cdot (tan^2 theta+1)$






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          Hint:



          Use the fact that



          $$tan^2 (theta)+1=sec^2(theta)$$



          Now , $sec^n theta=sec^{n-2} theta cdot sec^2 theta=sec^{n-2}theta cdot (tan^2 theta+1)$






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            Hint:



            Use the fact that



            $$tan^2 (theta)+1=sec^2(theta)$$



            Now , $sec^n theta=sec^{n-2} theta cdot sec^2 theta=sec^{n-2}theta cdot (tan^2 theta+1)$






            share|cite|improve this answer









            $endgroup$



            Hint:



            Use the fact that



            $$tan^2 (theta)+1=sec^2(theta)$$



            Now , $sec^n theta=sec^{n-2} theta cdot sec^2 theta=sec^{n-2}theta cdot (tan^2 theta+1)$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 9 '13 at 14:42









            InceptioInceptio

            7,0491535




            7,0491535























                0












                $begingroup$

                Hint:



                $$int{sec^{n}(theta)}dtheta= int{sec^{n-2}(theta)} sec^2(theta) dtheta$$



                Integrate by parts.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Hint:



                  $$int{sec^{n}(theta)}dtheta= int{sec^{n-2}(theta)} sec^2(theta) dtheta$$



                  Integrate by parts.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Hint:



                    $$int{sec^{n}(theta)}dtheta= int{sec^{n-2}(theta)} sec^2(theta) dtheta$$



                    Integrate by parts.






                    share|cite|improve this answer









                    $endgroup$



                    Hint:



                    $$int{sec^{n}(theta)}dtheta= int{sec^{n-2}(theta)} sec^2(theta) dtheta$$



                    Integrate by parts.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 9 '13 at 14:43









                    N. S.N. S.

                    102k6111208




                    102k6111208























                        0












                        $begingroup$

                        Hint:



                        remember that $d(tan theta) = frac{dtheta}{cos^2theta} = sec^2 theta dtheta$, integrate by parts, then collect the like terms with $int frac{dtheta}{cos^n theta}$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Hint:



                          remember that $d(tan theta) = frac{dtheta}{cos^2theta} = sec^2 theta dtheta$, integrate by parts, then collect the like terms with $int frac{dtheta}{cos^n theta}$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Hint:



                            remember that $d(tan theta) = frac{dtheta}{cos^2theta} = sec^2 theta dtheta$, integrate by parts, then collect the like terms with $int frac{dtheta}{cos^n theta}$






                            share|cite|improve this answer









                            $endgroup$



                            Hint:



                            remember that $d(tan theta) = frac{dtheta}{cos^2theta} = sec^2 theta dtheta$, integrate by parts, then collect the like terms with $int frac{dtheta}{cos^n theta}$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 9 '13 at 14:49









                            Doctor DanDoctor Dan

                            63137




                            63137






























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