Evaluate $intfrac {csc^2{x}-2005}{cos^{2005}{x}} dx $
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
edited Dec 28 '18 at 13:41
Eevee Trainer
5,0271734
asked Aug 1 '14 at 17:56
HenryHenry
2,21711249
add a comment |
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
edited Dec 28 '18 at 13:41
Eevee Trainer
5,0271734
asked Aug 1 '14 at 17:56
HenryHenry
2,21711249
Can you show the work which you've tried? and specifically where you got stuck when doing it?
– Rivasa
Aug 1 '14 at 17:58
add a comment |
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
edited Dec 28 '18 at 13:41
Eevee Trainer
5,0271734
asked Aug 1 '14 at 17:56
HenryHenry
2,21711249
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
edited Dec 28 '18 at 13:41
Eevee Trainer
5,0271734
asked Aug 1 '14 at 17:56
HenryHenry
2,21711249
edited Dec 28 '18 at 13:41
Eevee Trainer
5,0271734
asked Aug 1 '14 at 17:56
HenryHenry
2,21711249
edited Dec 28 '18 at 13:41
Eevee Trainer
5,0271734
edited Dec 28 '18 at 13:41
Eevee Trainer
5,0271734
edited Dec 28 '18 at 13:41
Eevee Trainer
5,0271734
5,0271734
asked Aug 1 '14 at 17:56
HenryHenry
2,21711249
asked Aug 1 '14 at 17:56
HenryHenry
2,21711249
asked Aug 1 '14 at 17:56
HenryHenry
2,21711249
2,21711249
Can you show the work which you've tried? and specifically where you got stuck when doing it?
– Rivasa
Aug 1 '14 at 17:58
add a comment |
Can you show the work which you've tried? and specifically where you got stuck when doing it?
– Rivasa
Aug 1 '14 at 17:58
Can you show the work which you've tried? and specifically where you got stuck when doing it?
– Rivasa
Aug 1 '14 at 17:58
Can you show the work which you've tried? and specifically where you got stuck when doing it?
– Rivasa
Aug 1 '14 at 17:58
add a comment |
1 Answer
1
active
oldest
votes
HINT:
Integrate by parts
$$int(sec^{2005}xcdotcsc^2x)dx$$
$$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$
answered Aug 1 '14 at 18:01
lab bhattacharjeelab bhattacharjee
223k15156274
2
Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
– Henry
Aug 1 '14 at 18:24
@Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
– Tunk-Fey
Aug 1 '14 at 18:38
@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?
– lab bhattacharjee
Aug 2 '14 at 4:06
@labbhattacharjee Oh right! Thanks a lot!
– Henry
Aug 2 '14 at 20:26
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
HINT:
Integrate by parts
$$int(sec^{2005}xcdotcsc^2x)dx$$
$$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$
answered Aug 1 '14 at 18:01
lab bhattacharjeelab bhattacharjee
223k15156274
2
Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
– Henry
Aug 1 '14 at 18:24
@Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
– Tunk-Fey
Aug 1 '14 at 18:38
@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?
– lab bhattacharjee
Aug 2 '14 at 4:06
@labbhattacharjee Oh right! Thanks a lot!
– Henry
Aug 2 '14 at 20:26
add a comment |
HINT:
Integrate by parts
$$int(sec^{2005}xcdotcsc^2x)dx$$
$$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$
answered Aug 1 '14 at 18:01
lab bhattacharjeelab bhattacharjee
223k15156274
2
Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
– Henry
Aug 1 '14 at 18:24
@Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
– Tunk-Fey
Aug 1 '14 at 18:38
@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?
– lab bhattacharjee
Aug 2 '14 at 4:06
@labbhattacharjee Oh right! Thanks a lot!
– Henry
Aug 2 '14 at 20:26
add a comment |
HINT:
Integrate by parts
$$int(sec^{2005}xcdotcsc^2x)dx$$
$$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$
answered Aug 1 '14 at 18:01
lab bhattacharjeelab bhattacharjee
223k15156274
HINT:
Integrate by parts
$$int(sec^{2005}xcdotcsc^2x)dx$$
$$=sec^{2005}xintcsc^2x dx-intleft(frac{d(sec^{2005}x)}{dx}intcsc^2x dxright)dx$$
answered Aug 1 '14 at 18:01
lab bhattacharjeelab bhattacharjee
223k15156274
answered Aug 1 '14 at 18:01
lab bhattacharjeelab bhattacharjee
223k15156274
answered Aug 1 '14 at 18:01
lab bhattacharjeelab bhattacharjee
223k15156274
answered Aug 1 '14 at 18:01
lab bhattacharjeelab bhattacharjee
223k15156274
223k15156274
2
Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
– Henry
Aug 1 '14 at 18:24
@Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
– Tunk-Fey
Aug 1 '14 at 18:38
@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?
– lab bhattacharjee
Aug 2 '14 at 4:06
@labbhattacharjee Oh right! Thanks a lot!
– Henry
Aug 2 '14 at 20:26
add a comment |
2
Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
– Henry
Aug 1 '14 at 18:24
@Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
– Tunk-Fey
Aug 1 '14 at 18:38
@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?
– lab bhattacharjee
Aug 2 '14 at 4:06
@labbhattacharjee Oh right! Thanks a lot!
– Henry
Aug 2 '14 at 20:26
2
2
Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
– Henry
Aug 1 '14 at 18:24
Are you talking about splitting it into two integrals? In that case how will you evaluate $-2005intsec^{2005}{x}dx$ ?
– Henry
Aug 1 '14 at 18:24
@Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
– Tunk-Fey
Aug 1 '14 at 18:38
@Samurai No need to evaluate it since it will be vanished by $2$nd integral in the RHS.
– Tunk-Fey
Aug 1 '14 at 18:38
@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?
– lab bhattacharjee
Aug 2 '14 at 4:06
@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?
– lab bhattacharjee
Aug 2 '14 at 4:06
@labbhattacharjee Oh right! Thanks a lot!
– Henry
Aug 2 '14 at 20:26
@labbhattacharjee Oh right! Thanks a lot!
– Henry
Aug 2 '14 at 20:26
add a comment |
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Can you show the work which you've tried? and specifically where you got stuck when doing it?
– Rivasa
Aug 1 '14 at 17:58