Evaluate $intfrac {csc^2{x}-2005}{cos^{2005}{x}} dx $












3












$begingroup$



Evaluate the indefinite integral
$$intfrac {csc^2{x}-2005}{cos^{2005}{x}} dx$$




I tried multiplying and dividing by $sec^2 {x} $ and then setting $tan{x}=y$ but no good. Then I set $cos {x}=t $ and tried to create $sin {x} $ in the numerator. But the integral which came was also a difficult one.



Please Help!



Thanks!



P.S. I am a high school student so kindly use elementary methods only. Thanks again!






calculus integration trigonometry indefinite-integrals







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  • $begingroup$
    Can you show the work which you've tried? and specifically where you got stuck when doing it?
    $endgroup$
    – Rivasa
    Aug 1 '14 at 17:58
















3












$begingroup$



Evaluate the indefinite integral
$$intfrac {csc^2{x}-2005}{cos^{2005}{x}} dx$$




I tried multiplying and dividing by $sec^2 {x} $ and then setting $tan{x}=y$ but no good. Then I set $cos {x}=t $ and tried to create $sin {x} $ in the numerator. But the integral which came was also a difficult one.



Please Help!



Thanks!



P.S. I am a high school student so kindly use elementary methods only. Thanks again!






calculus integration trigonometry indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you show the work which you've tried? and specifically where you got stuck when doing it?
    $endgroup$
    – Rivasa
    Aug 1 '14 at 17:58














3












3








3


4



$begingroup$



Evaluate the indefinite integral
$$intfrac {csc^2{x}-2005}{cos^{2005}{x}} dx$$




I tried multiplying and dividing by $sec^2 {x} $ and then setting $tan{x}=y$ but no good. Then I set $cos {x}=t $ and tried to create $sin {x} $ in the numerator. But the integral which came was also a difficult one.



Please Help!



Thanks!



P.S. I am a high school student so kindly use elementary methods only. Thanks again!






calculus integration trigonometry indefinite-integrals







share|cite|improve this question











$endgroup$





Evaluate the indefinite integral
$$intfrac {csc^2{x}-2005}{cos^{2005}{x}} dx$$




I tried multiplying and dividing by $sec^2 {x} $ and then setting $tan{x}=y$ but no good. Then I set $cos {x}=t $ and tried to create $sin {x} $ in the numerator. But the integral which came was also a difficult one.



Please Help!



Thanks!



P.S. I am a high school student so kindly use elementary methods only. Thanks again!






calculus integration trigonometry indefinite-integrals




calculus integration trigonometry indefinite-integrals






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share|cite|improve this question













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edited Dec 28 '18 at 13:41









Eevee Trainer

6,0931936




6,0931936










asked Aug 1 '14 at 17:56









HenryHenry

2,22211249




2,22211249












  • $begingroup$
    Can you show the work which you've tried? and specifically where you got stuck when doing it?
    $endgroup$
    – Rivasa
    Aug 1 '14 at 17:58


















  • $begingroup$
    Can you show the work which you've tried? and specifically where you got stuck when doing it?
    $endgroup$
    – Rivasa
    Aug 1 '14 at 17:58
















$begingroup$
Can you show the work which you've tried? and specifically where you got stuck when doing it?
$endgroup$
– Rivasa
Aug 1 '14 at 17:58




$begingroup$
Can you show the work which you've tried? and specifically where you got stuck when doing it?
$endgroup$
– Rivasa
Aug 1 '14 at 17:58










2 Answers
2






active

oldest

votes


















9












$begingroup$

HINT:



Integrate by parts
$$int(sec^{2005}xcdotcsc^2x)dx$$



$$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
    $endgroup$
    – Henry
    Aug 1 '14 at 18:24












  • $begingroup$
    @Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
    $endgroup$
    – Tunk-Fey
    Aug 1 '14 at 18:38












  • $begingroup$
    @Samurai, $$intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx=int2005sec^{2004}x(sec xtan x)(-cot x) dx=-2005intsec^{2005}x dx,$$ right?
    $endgroup$
    – lab bhattacharjee
    Aug 2 '14 at 4:06










  • $begingroup$
    @labbhattacharjee Oh right! Thanks a lot!
    $endgroup$
    – Henry
    Aug 2 '14 at 20:26



















1












$begingroup$

$$int frac{csc^2 x-2005}{cos^{2005}x}dx = int frac{cos^{2005}xcsc^2 x-2005cos^{2005}x}{(cos^{2005}x)^2}dx$$



$$int frac{d}{dx}bigg(frac{-cot x}{cos^{2005}x}bigg)dx = -frac{cot x}{cos^{2005}x}+C$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    HINT:



    Integrate by parts
    $$int(sec^{2005}xcdotcsc^2x)dx$$



    $$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
      $endgroup$
      – Henry
      Aug 1 '14 at 18:24












    • $begingroup$
      @Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
      $endgroup$
      – Tunk-Fey
      Aug 1 '14 at 18:38












    • $begingroup$
      @Samurai, $$intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx=int2005sec^{2004}x(sec xtan x)(-cot x) dx=-2005intsec^{2005}x dx,$$ right?
      $endgroup$
      – lab bhattacharjee
      Aug 2 '14 at 4:06










    • $begingroup$
      @labbhattacharjee Oh right! Thanks a lot!
      $endgroup$
      – Henry
      Aug 2 '14 at 20:26
















    9












    $begingroup$

    HINT:



    Integrate by parts
    $$int(sec^{2005}xcdotcsc^2x)dx$$



    $$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
      $endgroup$
      – Henry
      Aug 1 '14 at 18:24












    • $begingroup$
      @Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
      $endgroup$
      – Tunk-Fey
      Aug 1 '14 at 18:38












    • $begingroup$
      @Samurai, $$intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx=int2005sec^{2004}x(sec xtan x)(-cot x) dx=-2005intsec^{2005}x dx,$$ right?
      $endgroup$
      – lab bhattacharjee
      Aug 2 '14 at 4:06










    • $begingroup$
      @labbhattacharjee Oh right! Thanks a lot!
      $endgroup$
      – Henry
      Aug 2 '14 at 20:26














    9












    9








    9





    $begingroup$

    HINT:



    Integrate by parts
    $$int(sec^{2005}xcdotcsc^2x)dx$$



    $$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$






    share|cite|improve this answer









    $endgroup$



    HINT:



    Integrate by parts
    $$int(sec^{2005}xcdotcsc^2x)dx$$



    $$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 1 '14 at 18:01









    lab bhattacharjeelab bhattacharjee

    226k15157275




    226k15157275








    • 2




      $begingroup$
      Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
      $endgroup$
      – Henry
      Aug 1 '14 at 18:24












    • $begingroup$
      @Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
      $endgroup$
      – Tunk-Fey
      Aug 1 '14 at 18:38












    • $begingroup$
      @Samurai, $$intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx=int2005sec^{2004}x(sec xtan x)(-cot x) dx=-2005intsec^{2005}x dx,$$ right?
      $endgroup$
      – lab bhattacharjee
      Aug 2 '14 at 4:06










    • $begingroup$
      @labbhattacharjee Oh right! Thanks a lot!
      $endgroup$
      – Henry
      Aug 2 '14 at 20:26














    • 2




      $begingroup$
      Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
      $endgroup$
      – Henry
      Aug 1 '14 at 18:24












    • $begingroup$
      @Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
      $endgroup$
      – Tunk-Fey
      Aug 1 '14 at 18:38












    • $begingroup$
      @Samurai, $$intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx=int2005sec^{2004}x(sec xtan x)(-cot x) dx=-2005intsec^{2005}x dx,$$ right?
      $endgroup$
      – lab bhattacharjee
      Aug 2 '14 at 4:06










    • $begingroup$
      @labbhattacharjee Oh right! Thanks a lot!
      $endgroup$
      – Henry
      Aug 2 '14 at 20:26








    2




    2




    $begingroup$
    Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
    $endgroup$
    – Henry
    Aug 1 '14 at 18:24






    $begingroup$
    Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
    $endgroup$
    – Henry
    Aug 1 '14 at 18:24














    $begingroup$
    @Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
    $endgroup$
    – Tunk-Fey
    Aug 1 '14 at 18:38






    $begingroup$
    @Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
    $endgroup$
    – Tunk-Fey
    Aug 1 '14 at 18:38














    $begingroup$
    @Samurai, $$intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx=int2005sec^{2004}x(sec xtan x)(-cot x) dx=-2005intsec^{2005}x dx,$$ right?
    $endgroup$
    – lab bhattacharjee
    Aug 2 '14 at 4:06




    $begingroup$
    @Samurai, $$intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx=int2005sec^{2004}x(sec xtan x)(-cot x) dx=-2005intsec^{2005}x dx,$$ right?
    $endgroup$
    – lab bhattacharjee
    Aug 2 '14 at 4:06












    $begingroup$
    @labbhattacharjee Oh right! Thanks a lot!
    $endgroup$
    – Henry
    Aug 2 '14 at 20:26




    $begingroup$
    @labbhattacharjee Oh right! Thanks a lot!
    $endgroup$
    – Henry
    Aug 2 '14 at 20:26











    1












    $begingroup$

    $$int frac{csc^2 x-2005}{cos^{2005}x}dx = int frac{cos^{2005}xcsc^2 x-2005cos^{2005}x}{(cos^{2005}x)^2}dx$$



    $$int frac{d}{dx}bigg(frac{-cot x}{cos^{2005}x}bigg)dx = -frac{cot x}{cos^{2005}x}+C$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$int frac{csc^2 x-2005}{cos^{2005}x}dx = int frac{cos^{2005}xcsc^2 x-2005cos^{2005}x}{(cos^{2005}x)^2}dx$$



      $$int frac{d}{dx}bigg(frac{-cot x}{cos^{2005}x}bigg)dx = -frac{cot x}{cos^{2005}x}+C$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$int frac{csc^2 x-2005}{cos^{2005}x}dx = int frac{cos^{2005}xcsc^2 x-2005cos^{2005}x}{(cos^{2005}x)^2}dx$$



        $$int frac{d}{dx}bigg(frac{-cot x}{cos^{2005}x}bigg)dx = -frac{cot x}{cos^{2005}x}+C$$






        share|cite|improve this answer









        $endgroup$



        $$int frac{csc^2 x-2005}{cos^{2005}x}dx = int frac{cos^{2005}xcsc^2 x-2005cos^{2005}x}{(cos^{2005}x)^2}dx$$



        $$int frac{d}{dx}bigg(frac{-cot x}{cos^{2005}x}bigg)dx = -frac{cot x}{cos^{2005}x}+C$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 5 at 13:29









        jackyjacky

        861612




        861612






























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