Why Doesn't my Integration to Square Root Tan(x) Work?
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What is wrong with this solution?
https://www.mathcha.io/editor/egPXHEXSj3CD1ip1
$$I=int sqrt{tan x}dx$$
$u=sqrt{tan x}Rightarrow frac{2u}{u^2+1}du=dx$:
$$I=2intfrac{u^2}{u^2+1}du$$
$$I=2intfrac{u^2+1-1}{u^2+1}du$$
$$I=2bigg(int 1du-intfrac1{u^2+1}dubigg)$$
$$I=2big(u-arctan ubig)+C$$
$$I=2sqrt{tan x}-2arctansqrt{tan x}+C$$
Thanks!
Please do not link other solutions of this integral, I just want to know why the one I did is incorrect, thanks!
integration
edited Nov 24 at 1:53
clathratus
2,673325
asked Nov 24 at 0:50
Zach
12716
add a comment |
up vote
-1
down vote
favorite
What is wrong with this solution?
https://www.mathcha.io/editor/egPXHEXSj3CD1ip1
$$I=int sqrt{tan x}dx$$
$u=sqrt{tan x}Rightarrow frac{2u}{u^2+1}du=dx$:
$$I=2intfrac{u^2}{u^2+1}du$$
$$I=2intfrac{u^2+1-1}{u^2+1}du$$
$$I=2bigg(int 1du-intfrac1{u^2+1}dubigg)$$
$$I=2big(u-arctan ubig)+C$$
$$I=2sqrt{tan x}-2arctansqrt{tan x}+C$$
Thanks!
Please do not link other solutions of this integral, I just want to know why the one I did is incorrect, thanks!
integration
edited Nov 24 at 1:53
clathratus
2,673325
asked Nov 24 at 0:50
Zach
12716
1
Figuring out LaTeX is pretty much a requirement for using this site. Plus, you'll gain a great skill if you ever have to write-up math in the future.
– JonathanZ
Nov 24 at 1:09
1
I added the LaTex as an edit. Take a look at it so you can see how it's done.
– clathratus
Nov 24 at 1:16
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
What is wrong with this solution?
https://www.mathcha.io/editor/egPXHEXSj3CD1ip1
$$I=int sqrt{tan x}dx$$
$u=sqrt{tan x}Rightarrow frac{2u}{u^2+1}du=dx$:
$$I=2intfrac{u^2}{u^2+1}du$$
$$I=2intfrac{u^2+1-1}{u^2+1}du$$
$$I=2bigg(int 1du-intfrac1{u^2+1}dubigg)$$
$$I=2big(u-arctan ubig)+C$$
$$I=2sqrt{tan x}-2arctansqrt{tan x}+C$$
Thanks!
Please do not link other solutions of this integral, I just want to know why the one I did is incorrect, thanks!
integration
edited Nov 24 at 1:53
clathratus
2,673325
asked Nov 24 at 0:50
Zach
12716
What is wrong with this solution?
https://www.mathcha.io/editor/egPXHEXSj3CD1ip1
$$I=int sqrt{tan x}dx$$
$u=sqrt{tan x}Rightarrow frac{2u}{u^2+1}du=dx$:
$$I=2intfrac{u^2}{u^2+1}du$$
$$I=2intfrac{u^2+1-1}{u^2+1}du$$
$$I=2bigg(int 1du-intfrac1{u^2+1}dubigg)$$
$$I=2big(u-arctan ubig)+C$$
$$I=2sqrt{tan x}-2arctansqrt{tan x}+C$$
Thanks!
Please do not link other solutions of this integral, I just want to know why the one I did is incorrect, thanks!
integration
integration
edited Nov 24 at 1:53
clathratus
2,673325
asked Nov 24 at 0:50
Zach
12716
edited Nov 24 at 1:53
clathratus
2,673325
asked Nov 24 at 0:50
Zach
12716
edited Nov 24 at 1:53
clathratus
2,673325
edited Nov 24 at 1:53
clathratus
2,673325
edited Nov 24 at 1:53
clathratus
2,673325
2,673325
asked Nov 24 at 0:50
Zach
12716
asked Nov 24 at 0:50
Zach
12716
asked Nov 24 at 0:50
Zach
12716
12716
1
Figuring out LaTeX is pretty much a requirement for using this site. Plus, you'll gain a great skill if you ever have to write-up math in the future.
– JonathanZ
Nov 24 at 1:09
1
I added the LaTex as an edit. Take a look at it so you can see how it's done.
– clathratus
Nov 24 at 1:16
add a comment |
1
Figuring out LaTeX is pretty much a requirement for using this site. Plus, you'll gain a great skill if you ever have to write-up math in the future.
– JonathanZ
Nov 24 at 1:09
1
I added the LaTex as an edit. Take a look at it so you can see how it's done.
– clathratus
Nov 24 at 1:16
1
1
Figuring out LaTeX is pretty much a requirement for using this site. Plus, you'll gain a great skill if you ever have to write-up math in the future.
– JonathanZ
Nov 24 at 1:09
Figuring out LaTeX is pretty much a requirement for using this site. Plus, you'll gain a great skill if you ever have to write-up math in the future.
– JonathanZ
Nov 24 at 1:09
1
1
I added the LaTex as an edit. Take a look at it so you can see how it's done.
– clathratus
Nov 24 at 1:16
I added the LaTex as an edit. Take a look at it so you can see how it's done.
– clathratus
Nov 24 at 1:16
add a comment |
2 Answers
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$cos^2 x neq frac{1}{1+(sqrt{tan{x}})^2}$. $cos^2 x = frac{1}{1+(sqrt{tan{x}})^4} = frac{1}{1+u^4}$.
answered Nov 24 at 0:58
Eric Towers
31.6k22265
add a comment |
up vote
0
down vote
$$u=sqrt{tan x}implies x= tan ^{-1}left(u^2right)implies dx=frac{2 u}{u^4+1},du$$
$$I=int sqrt{tan x},dx=2int frac{ u^2}{u^4+1},du=int frac{ u^2}{left(u^2-sqrt{2} u+1right) left(u^2+sqrt{2} u+1right)},du$$ Now, partial fraction decomposition
$$I=frac 1 {sqrt 2}left(int frac{u}{u^2-sqrt{2} u+1} ,du-int frac{u}{u^2+sqrt{2} u+1},du right)$$ which becomes to be quite easy.
answered Nov 24 at 5:34
Claude Leibovici
118k1156131
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
$cos^2 x neq frac{1}{1+(sqrt{tan{x}})^2}$. $cos^2 x = frac{1}{1+(sqrt{tan{x}})^4} = frac{1}{1+u^4}$.
answered Nov 24 at 0:58
Eric Towers
31.6k22265
add a comment |
up vote
4
down vote
$cos^2 x neq frac{1}{1+(sqrt{tan{x}})^2}$. $cos^2 x = frac{1}{1+(sqrt{tan{x}})^4} = frac{1}{1+u^4}$.
answered Nov 24 at 0:58
Eric Towers
31.6k22265
add a comment |
up vote
4
down vote
up vote
4
down vote
$cos^2 x neq frac{1}{1+(sqrt{tan{x}})^2}$. $cos^2 x = frac{1}{1+(sqrt{tan{x}})^4} = frac{1}{1+u^4}$.
answered Nov 24 at 0:58
Eric Towers
31.6k22265
$cos^2 x neq frac{1}{1+(sqrt{tan{x}})^2}$. $cos^2 x = frac{1}{1+(sqrt{tan{x}})^4} = frac{1}{1+u^4}$.
answered Nov 24 at 0:58
Eric Towers
31.6k22265
answered Nov 24 at 0:58
Eric Towers
31.6k22265
answered Nov 24 at 0:58
Eric Towers
31.6k22265
answered Nov 24 at 0:58
Eric Towers
31.6k22265
31.6k22265
add a comment |
add a comment |
up vote
0
down vote
$$u=sqrt{tan x}implies x= tan ^{-1}left(u^2right)implies dx=frac{2 u}{u^4+1},du$$
$$I=int sqrt{tan x},dx=2int frac{ u^2}{u^4+1},du=int frac{ u^2}{left(u^2-sqrt{2} u+1right) left(u^2+sqrt{2} u+1right)},du$$ Now, partial fraction decomposition
$$I=frac 1 {sqrt 2}left(int frac{u}{u^2-sqrt{2} u+1} ,du-int frac{u}{u^2+sqrt{2} u+1},du right)$$ which becomes to be quite easy.
answered Nov 24 at 5:34
Claude Leibovici
118k1156131
add a comment |
up vote
0
down vote
$$u=sqrt{tan x}implies x= tan ^{-1}left(u^2right)implies dx=frac{2 u}{u^4+1},du$$
$$I=int sqrt{tan x},dx=2int frac{ u^2}{u^4+1},du=int frac{ u^2}{left(u^2-sqrt{2} u+1right) left(u^2+sqrt{2} u+1right)},du$$ Now, partial fraction decomposition
$$I=frac 1 {sqrt 2}left(int frac{u}{u^2-sqrt{2} u+1} ,du-int frac{u}{u^2+sqrt{2} u+1},du right)$$ which becomes to be quite easy.
answered Nov 24 at 5:34
Claude Leibovici
118k1156131
add a comment |
up vote
0
down vote
up vote
0
down vote
$$u=sqrt{tan x}implies x= tan ^{-1}left(u^2right)implies dx=frac{2 u}{u^4+1},du$$
$$I=int sqrt{tan x},dx=2int frac{ u^2}{u^4+1},du=int frac{ u^2}{left(u^2-sqrt{2} u+1right) left(u^2+sqrt{2} u+1right)},du$$ Now, partial fraction decomposition
$$I=frac 1 {sqrt 2}left(int frac{u}{u^2-sqrt{2} u+1} ,du-int frac{u}{u^2+sqrt{2} u+1},du right)$$ which becomes to be quite easy.
answered Nov 24 at 5:34
Claude Leibovici
118k1156131
$$u=sqrt{tan x}implies x= tan ^{-1}left(u^2right)implies dx=frac{2 u}{u^4+1},du$$
$$I=int sqrt{tan x},dx=2int frac{ u^2}{u^4+1},du=int frac{ u^2}{left(u^2-sqrt{2} u+1right) left(u^2+sqrt{2} u+1right)},du$$ Now, partial fraction decomposition
$$I=frac 1 {sqrt 2}left(int frac{u}{u^2-sqrt{2} u+1} ,du-int frac{u}{u^2+sqrt{2} u+1},du right)$$ which becomes to be quite easy.
answered Nov 24 at 5:34
Claude Leibovici
118k1156131
answered Nov 24 at 5:34
Claude Leibovici
118k1156131
answered Nov 24 at 5:34
Claude Leibovici
118k1156131
answered Nov 24 at 5:34
Claude Leibovici
118k1156131
118k1156131
add a comment |
add a comment |
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1
Figuring out LaTeX is pretty much a requirement for using this site. Plus, you'll gain a great skill if you ever have to write-up math in the future.
– JonathanZ
Nov 24 at 1:09
1
I added the LaTex as an edit. Take a look at it so you can see how it's done.
– clathratus
Nov 24 at 1:16