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How to simplify $intsqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1dx$?
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up vote
18
down vote
favorite
I have been asked about the following integral:
$$intsqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1dx$$
I think this is a joke of bad taste. I have tried every elementary method of integration which i know, also i tried integrating using Maple but as i suspected, the integrad doesn't have an anti derivative. Any ideas?
real-analysis integration indefinite-integrals radicals nested-radicals
edited Aug 1 at 7:38
user21820
35.8k440136
asked Aug 1 at 4:23
Johnny chad
1106
add a comment |Â
up vote
18
down vote
favorite
I have been asked about the following integral:
$$intsqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1dx$$
I think this is a joke of bad taste. I have tried every elementary method of integration which i know, also i tried integrating using Maple but as i suspected, the integrad doesn't have an anti derivative. Any ideas?
real-analysis integration indefinite-integrals radicals nested-radicals
edited Aug 1 at 7:38
user21820
35.8k440136
asked Aug 1 at 4:23
Johnny chad
1106
You could always put it into WolframAlpha, see if you get anything remotely nice and if so reverse engineer a solution from there. But be warned: some times WA gives ugly answers when there exist nice ones.
– Arthur
Aug 1 at 4:33
WA doesn't give anything nice back: wolframalpha.com/input/…
– Mason
Aug 1 at 4:36
3
All of these kinds of integrals are jokes in bad taste. The universe is laughing at the limited expressive strength of our mathematics.
– Daniel
Aug 1 at 6:32
3
Can't the title be more descriptive? .........
– user202729
Aug 1 at 7:21
@Mason: WA often gives Wrong Answers, so even if it says something you still have to know how to verify it.
– user21820
Aug 1 at 7:40
add a comment |Â
up vote
18
down vote
favorite
up vote
18
down vote
favorite
I have been asked about the following integral:
$$intsqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1dx$$
I think this is a joke of bad taste. I have tried every elementary method of integration which i know, also i tried integrating using Maple but as i suspected, the integrad doesn't have an anti derivative. Any ideas?
real-analysis integration indefinite-integrals radicals nested-radicals
edited Aug 1 at 7:38
user21820
35.8k440136
asked Aug 1 at 4:23
Johnny chad
1106
I have been asked about the following integral:
$$intsqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1dx$$
I think this is a joke of bad taste. I have tried every elementary method of integration which i know, also i tried integrating using Maple but as i suspected, the integrad doesn't have an anti derivative. Any ideas?
real-analysis integration indefinite-integrals radicals nested-radicals
edited Aug 1 at 7:38
user21820
35.8k440136
asked Aug 1 at 4:23
Johnny chad
1106
edited Aug 1 at 7:38
user21820
35.8k440136
edited Aug 1 at 7:38
user21820
35.8k440136
edited Aug 1 at 7:38
user21820
35.8k440136
35.8k440136
asked Aug 1 at 4:23
Johnny chad
1106
asked Aug 1 at 4:23
Johnny chad
1106
asked Aug 1 at 4:23
Johnny chad
1106
1106
You could always put it into WolframAlpha, see if you get anything remotely nice and if so reverse engineer a solution from there. But be warned: some times WA gives ugly answers when there exist nice ones.
– Arthur
Aug 1 at 4:33
WA doesn't give anything nice back: wolframalpha.com/input/…
– Mason
Aug 1 at 4:36
3
All of these kinds of integrals are jokes in bad taste. The universe is laughing at the limited expressive strength of our mathematics.
– Daniel
Aug 1 at 6:32
3
Can't the title be more descriptive? .........
– user202729
Aug 1 at 7:21
@Mason: WA often gives Wrong Answers, so even if it says something you still have to know how to verify it.
– user21820
Aug 1 at 7:40
add a comment |Â
You could always put it into WolframAlpha, see if you get anything remotely nice and if so reverse engineer a solution from there. But be warned: some times WA gives ugly answers when there exist nice ones.
– Arthur
Aug 1 at 4:33
WA doesn't give anything nice back: wolframalpha.com/input/…
– Mason
Aug 1 at 4:36
3
All of these kinds of integrals are jokes in bad taste. The universe is laughing at the limited expressive strength of our mathematics.
– Daniel
Aug 1 at 6:32
3
Can't the title be more descriptive? .........
– user202729
Aug 1 at 7:21
@Mason: WA often gives Wrong Answers, so even if it says something you still have to know how to verify it.
– user21820
Aug 1 at 7:40
You could always put it into WolframAlpha, see if you get anything remotely nice and if so reverse engineer a solution from there. But be warned: some times WA gives ugly answers when there exist nice ones.
– Arthur
Aug 1 at 4:33
You could always put it into WolframAlpha, see if you get anything remotely nice and if so reverse engineer a solution from there. But be warned: some times WA gives ugly answers when there exist nice ones.
– Arthur
Aug 1 at 4:33
WA doesn't give anything nice back: wolframalpha.com/input/…
– Mason
Aug 1 at 4:36
WA doesn't give anything nice back: wolframalpha.com/input/…
– Mason
Aug 1 at 4:36
3
3
All of these kinds of integrals are jokes in bad taste. The universe is laughing at the limited expressive strength of our mathematics.
– Daniel
Aug 1 at 6:32
All of these kinds of integrals are jokes in bad taste. The universe is laughing at the limited expressive strength of our mathematics.
– Daniel
Aug 1 at 6:32
3
3
Can't the title be more descriptive? .........
– user202729
Aug 1 at 7:21
Can't the title be more descriptive? .........
– user202729
Aug 1 at 7:21
@Mason: WA often gives Wrong Answers, so even if it says something you still have to know how to verify it.
– user21820
Aug 1 at 7:40
@Mason: WA often gives Wrong Answers, so even if it says something you still have to know how to verify it.
– user21820
Aug 1 at 7:40
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
41
down vote
accepted
Use $$sqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=left|x+sqrtx^2-1right|.$$
edited Aug 2 at 12:02
Kitegi
5511919
answered Aug 1 at 4:34
Michael Rozenberg
87.4k1577179
2
How did you discover that?
– Arthur
Aug 1 at 4:47
6
@Arthur I tried to get $(a+b)^4$ under the radical and it turns out.
– Michael Rozenberg
Aug 1 at 4:50
3
Desmos confirms that the two functions are the same.
– Toby Mak
Aug 1 at 4:57
It is a single hyperbola like curve with two disconnected parts. Each part can be obtained by changing sign in front of $sqrt...$
– Narasimham
Aug 1 at 5:13
What made you think of that? Pls I need to know.
– William
Aug 2 at 10:24
 |Â
show 1 more comment
up vote
8
down vote
Hint:
$(x+sqrtx^2-1)^4 = x^4+4x^3sqrtx^2-1+6x^2(x^2-1)+4x(x^2-1)sqrtx^2-1+(x^2-1)^2 = 8x^4-8x^2+1-4xsqrtx^2-1+8x^3sqrtx^2-1$
answered Aug 1 at 4:49
A. Pongrácz
1,304115
add a comment |Â
up vote
5
down vote
$$frac116left( sqrtx-1+sqrtx+1 right)^8=1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1$$
So the integral becomes
$$beginalign
& =intsqrt[4]frac116left( sqrtx-1+sqrtx+1 right)^8dx \
& =frac12intleft( sqrtx-1+sqrtx+1 right)^2dx \
& =intleft( x+sqrtx^2-1 right)dx \
endalign$$
answered Aug 1 at 8:09
Vincent Law
1,192110
7
While this is nice, I want to warn people that $$f(x):=frac12left(sqrtx-1+sqrtx+1right)^2$$ is not the same as $$g(x):=big|x+sqrtx^2-1big|,.$$ Over the reals, $f(x)$ is defined only for $xgeq 1$, but $g(x)$ is defined for all $x$ with $|x|geq 1$. I subtracted points in the past from students who had written something like this in their homework or exam papers. (But don't worry, I am not downvoting you.)
– Batominovski
Aug 1 at 8:55
add a comment |Â
up vote
4
down vote
Hint:
Let $textarcsecx=t, x=sec t$
Using Principal values, $0le tlepi,tnedfracpi2$
$sin t=sqrtleft(1-dfrac1xright)^2=dfracsqrtx^2-1$ as $sin tge0$
$tan t=sin tsec t=?$
$$1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=dfraccos^4t-8cos^2t+8-4sin tcos^2t+8sin tcos^4t$$
Now $cos^4t-8cos^2t+8-4sin tcos^2t+8sin t$
$=(1-sin^2t)^2-8(1-sin^2t)+8-4sin t(1-sin^2t)+8sin t$
$=sin^4t+4sin^3t+6sin^2t+4sin t+1$
$=(sin t+1)^4$
answered Aug 2 at 8:53
lab bhattacharjee
214k14152263
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
41
down vote
accepted
Use $$sqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=left|x+sqrtx^2-1right|.$$
edited Aug 2 at 12:02
Kitegi
5511919
answered Aug 1 at 4:34
Michael Rozenberg
87.4k1577179
2
How did you discover that?
– Arthur
Aug 1 at 4:47
6
@Arthur I tried to get $(a+b)^4$ under the radical and it turns out.
– Michael Rozenberg
Aug 1 at 4:50
3
Desmos confirms that the two functions are the same.
– Toby Mak
Aug 1 at 4:57
It is a single hyperbola like curve with two disconnected parts. Each part can be obtained by changing sign in front of $sqrt...$
– Narasimham
Aug 1 at 5:13
What made you think of that? Pls I need to know.
– William
Aug 2 at 10:24
 |Â
show 1 more comment
up vote
41
down vote
accepted
Use $$sqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=left|x+sqrtx^2-1right|.$$
edited Aug 2 at 12:02
Kitegi
5511919
answered Aug 1 at 4:34
Michael Rozenberg
87.4k1577179
2
How did you discover that?
– Arthur
Aug 1 at 4:47
6
@Arthur I tried to get $(a+b)^4$ under the radical and it turns out.
– Michael Rozenberg
Aug 1 at 4:50
3
Desmos confirms that the two functions are the same.
– Toby Mak
Aug 1 at 4:57
It is a single hyperbola like curve with two disconnected parts. Each part can be obtained by changing sign in front of $sqrt...$
– Narasimham
Aug 1 at 5:13
What made you think of that? Pls I need to know.
– William
Aug 2 at 10:24
 |Â
show 1 more comment
up vote
41
down vote
accepted
up vote
41
down vote
accepted
Use $$sqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=left|x+sqrtx^2-1right|.$$
edited Aug 2 at 12:02
Kitegi
5511919
answered Aug 1 at 4:34
Michael Rozenberg
87.4k1577179
Use $$sqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=left|x+sqrtx^2-1right|.$$
edited Aug 2 at 12:02
Kitegi
5511919
answered Aug 1 at 4:34
Michael Rozenberg
87.4k1577179
edited Aug 2 at 12:02
Kitegi
5511919
edited Aug 2 at 12:02
Kitegi
5511919
edited Aug 2 at 12:02
Kitegi
5511919
5511919
answered Aug 1 at 4:34
Michael Rozenberg
87.4k1577179
answered Aug 1 at 4:34
Michael Rozenberg
87.4k1577179
answered Aug 1 at 4:34
Michael Rozenberg
87.4k1577179
87.4k1577179
2
How did you discover that?
– Arthur
Aug 1 at 4:47
6
@Arthur I tried to get $(a+b)^4$ under the radical and it turns out.
– Michael Rozenberg
Aug 1 at 4:50
3
Desmos confirms that the two functions are the same.
– Toby Mak
Aug 1 at 4:57
It is a single hyperbola like curve with two disconnected parts. Each part can be obtained by changing sign in front of $sqrt...$
– Narasimham
Aug 1 at 5:13
What made you think of that? Pls I need to know.
– William
Aug 2 at 10:24
 |Â
show 1 more comment
2
How did you discover that?
– Arthur
Aug 1 at 4:47
6
@Arthur I tried to get $(a+b)^4$ under the radical and it turns out.
– Michael Rozenberg
Aug 1 at 4:50
3
Desmos confirms that the two functions are the same.
– Toby Mak
Aug 1 at 4:57
It is a single hyperbola like curve with two disconnected parts. Each part can be obtained by changing sign in front of $sqrt...$
– Narasimham
Aug 1 at 5:13
What made you think of that? Pls I need to know.
– William
Aug 2 at 10:24
2
2
How did you discover that?
– Arthur
Aug 1 at 4:47
How did you discover that?
– Arthur
Aug 1 at 4:47
6
6
@Arthur I tried to get $(a+b)^4$ under the radical and it turns out.
– Michael Rozenberg
Aug 1 at 4:50
@Arthur I tried to get $(a+b)^4$ under the radical and it turns out.
– Michael Rozenberg
Aug 1 at 4:50
3
3
Desmos confirms that the two functions are the same.
– Toby Mak
Aug 1 at 4:57
Desmos confirms that the two functions are the same.
– Toby Mak
Aug 1 at 4:57
It is a single hyperbola like curve with two disconnected parts. Each part can be obtained by changing sign in front of $sqrt...$
– Narasimham
Aug 1 at 5:13
It is a single hyperbola like curve with two disconnected parts. Each part can be obtained by changing sign in front of $sqrt...$
– Narasimham
Aug 1 at 5:13
What made you think of that? Pls I need to know.
– William
Aug 2 at 10:24
What made you think of that? Pls I need to know.
– William
Aug 2 at 10:24
 |Â
show 1 more comment
up vote
8
down vote
Hint:
$(x+sqrtx^2-1)^4 = x^4+4x^3sqrtx^2-1+6x^2(x^2-1)+4x(x^2-1)sqrtx^2-1+(x^2-1)^2 = 8x^4-8x^2+1-4xsqrtx^2-1+8x^3sqrtx^2-1$
answered Aug 1 at 4:49
A. Pongrácz
1,304115
add a comment |Â
up vote
8
down vote
Hint:
$(x+sqrtx^2-1)^4 = x^4+4x^3sqrtx^2-1+6x^2(x^2-1)+4x(x^2-1)sqrtx^2-1+(x^2-1)^2 = 8x^4-8x^2+1-4xsqrtx^2-1+8x^3sqrtx^2-1$
answered Aug 1 at 4:49
A. Pongrácz
1,304115
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Hint:
$(x+sqrtx^2-1)^4 = x^4+4x^3sqrtx^2-1+6x^2(x^2-1)+4x(x^2-1)sqrtx^2-1+(x^2-1)^2 = 8x^4-8x^2+1-4xsqrtx^2-1+8x^3sqrtx^2-1$
answered Aug 1 at 4:49
A. Pongrácz
1,304115
Hint:
$(x+sqrtx^2-1)^4 = x^4+4x^3sqrtx^2-1+6x^2(x^2-1)+4x(x^2-1)sqrtx^2-1+(x^2-1)^2 = 8x^4-8x^2+1-4xsqrtx^2-1+8x^3sqrtx^2-1$
answered Aug 1 at 4:49
A. Pongrácz
1,304115
answered Aug 1 at 4:49
A. Pongrácz
1,304115
answered Aug 1 at 4:49
A. Pongrácz
1,304115
answered Aug 1 at 4:49
A. Pongrácz
1,304115
1,304115
add a comment |Â
add a comment |Â
up vote
5
down vote
$$frac116left( sqrtx-1+sqrtx+1 right)^8=1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1$$
So the integral becomes
$$beginalign
& =intsqrt[4]frac116left( sqrtx-1+sqrtx+1 right)^8dx \
& =frac12intleft( sqrtx-1+sqrtx+1 right)^2dx \
& =intleft( x+sqrtx^2-1 right)dx \
endalign$$
answered Aug 1 at 8:09
Vincent Law
1,192110
7
While this is nice, I want to warn people that $$f(x):=frac12left(sqrtx-1+sqrtx+1right)^2$$ is not the same as $$g(x):=big|x+sqrtx^2-1big|,.$$ Over the reals, $f(x)$ is defined only for $xgeq 1$, but $g(x)$ is defined for all $x$ with $|x|geq 1$. I subtracted points in the past from students who had written something like this in their homework or exam papers. (But don't worry, I am not downvoting you.)
– Batominovski
Aug 1 at 8:55
add a comment |Â
up vote
5
down vote
$$frac116left( sqrtx-1+sqrtx+1 right)^8=1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1$$
So the integral becomes
$$beginalign
& =intsqrt[4]frac116left( sqrtx-1+sqrtx+1 right)^8dx \
& =frac12intleft( sqrtx-1+sqrtx+1 right)^2dx \
& =intleft( x+sqrtx^2-1 right)dx \
endalign$$
answered Aug 1 at 8:09
Vincent Law
1,192110
7
While this is nice, I want to warn people that $$f(x):=frac12left(sqrtx-1+sqrtx+1right)^2$$ is not the same as $$g(x):=big|x+sqrtx^2-1big|,.$$ Over the reals, $f(x)$ is defined only for $xgeq 1$, but $g(x)$ is defined for all $x$ with $|x|geq 1$. I subtracted points in the past from students who had written something like this in their homework or exam papers. (But don't worry, I am not downvoting you.)
– Batominovski
Aug 1 at 8:55
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$$frac116left( sqrtx-1+sqrtx+1 right)^8=1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1$$
So the integral becomes
$$beginalign
& =intsqrt[4]frac116left( sqrtx-1+sqrtx+1 right)^8dx \
& =frac12intleft( sqrtx-1+sqrtx+1 right)^2dx \
& =intleft( x+sqrtx^2-1 right)dx \
endalign$$
answered Aug 1 at 8:09
Vincent Law
1,192110
$$frac116left( sqrtx-1+sqrtx+1 right)^8=1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1$$
So the integral becomes
$$beginalign
& =intsqrt[4]frac116left( sqrtx-1+sqrtx+1 right)^8dx \
& =frac12intleft( sqrtx-1+sqrtx+1 right)^2dx \
& =intleft( x+sqrtx^2-1 right)dx \
endalign$$
answered Aug 1 at 8:09
Vincent Law
1,192110
edited Aug 1 at 8:34
answered Aug 1 at 8:09
Vincent Law
1,192110
answered Aug 1 at 8:09
Vincent Law
1,192110
answered Aug 1 at 8:09
Vincent Law
1,192110
1,192110
7
While this is nice, I want to warn people that $$f(x):=frac12left(sqrtx-1+sqrtx+1right)^2$$ is not the same as $$g(x):=big|x+sqrtx^2-1big|,.$$ Over the reals, $f(x)$ is defined only for $xgeq 1$, but $g(x)$ is defined for all $x$ with $|x|geq 1$. I subtracted points in the past from students who had written something like this in their homework or exam papers. (But don't worry, I am not downvoting you.)
– Batominovski
Aug 1 at 8:55
add a comment |Â
7
While this is nice, I want to warn people that $$f(x):=frac12left(sqrtx-1+sqrtx+1right)^2$$ is not the same as $$g(x):=big|x+sqrtx^2-1big|,.$$ Over the reals, $f(x)$ is defined only for $xgeq 1$, but $g(x)$ is defined for all $x$ with $|x|geq 1$. I subtracted points in the past from students who had written something like this in their homework or exam papers. (But don't worry, I am not downvoting you.)
– Batominovski
Aug 1 at 8:55
7
7
While this is nice, I want to warn people that $$f(x):=frac12left(sqrtx-1+sqrtx+1right)^2$$ is not the same as $$g(x):=big|x+sqrtx^2-1big|,.$$ Over the reals, $f(x)$ is defined only for $xgeq 1$, but $g(x)$ is defined for all $x$ with $|x|geq 1$. I subtracted points in the past from students who had written something like this in their homework or exam papers. (But don't worry, I am not downvoting you.)
– Batominovski
Aug 1 at 8:55
While this is nice, I want to warn people that $$f(x):=frac12left(sqrtx-1+sqrtx+1right)^2$$ is not the same as $$g(x):=big|x+sqrtx^2-1big|,.$$ Over the reals, $f(x)$ is defined only for $xgeq 1$, but $g(x)$ is defined for all $x$ with $|x|geq 1$. I subtracted points in the past from students who had written something like this in their homework or exam papers. (But don't worry, I am not downvoting you.)
– Batominovski
Aug 1 at 8:55
add a comment |Â
up vote
4
down vote
Hint:
Let $textarcsecx=t, x=sec t$
Using Principal values, $0le tlepi,tnedfracpi2$
$sin t=sqrtleft(1-dfrac1xright)^2=dfracsqrtx^2-1$ as $sin tge0$
$tan t=sin tsec t=?$
$$1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=dfraccos^4t-8cos^2t+8-4sin tcos^2t+8sin tcos^4t$$
Now $cos^4t-8cos^2t+8-4sin tcos^2t+8sin t$
$=(1-sin^2t)^2-8(1-sin^2t)+8-4sin t(1-sin^2t)+8sin t$
$=sin^4t+4sin^3t+6sin^2t+4sin t+1$
$=(sin t+1)^4$
answered Aug 2 at 8:53
lab bhattacharjee
214k14152263
add a comment |Â
up vote
4
down vote
Hint:
Let $textarcsecx=t, x=sec t$
Using Principal values, $0le tlepi,tnedfracpi2$
$sin t=sqrtleft(1-dfrac1xright)^2=dfracsqrtx^2-1$ as $sin tge0$
$tan t=sin tsec t=?$
$$1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=dfraccos^4t-8cos^2t+8-4sin tcos^2t+8sin tcos^4t$$
Now $cos^4t-8cos^2t+8-4sin tcos^2t+8sin t$
$=(1-sin^2t)^2-8(1-sin^2t)+8-4sin t(1-sin^2t)+8sin t$
$=sin^4t+4sin^3t+6sin^2t+4sin t+1$
$=(sin t+1)^4$
answered Aug 2 at 8:53
lab bhattacharjee
214k14152263
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Hint:
Let $textarcsecx=t, x=sec t$
Using Principal values, $0le tlepi,tnedfracpi2$
$sin t=sqrtleft(1-dfrac1xright)^2=dfracsqrtx^2-1$ as $sin tge0$
$tan t=sin tsec t=?$
$$1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=dfraccos^4t-8cos^2t+8-4sin tcos^2t+8sin tcos^4t$$
Now $cos^4t-8cos^2t+8-4sin tcos^2t+8sin t$
$=(1-sin^2t)^2-8(1-sin^2t)+8-4sin t(1-sin^2t)+8sin t$
$=sin^4t+4sin^3t+6sin^2t+4sin t+1$
$=(sin t+1)^4$
answered Aug 2 at 8:53
lab bhattacharjee
214k14152263
Hint:
Let $textarcsecx=t, x=sec t$
Using Principal values, $0le tlepi,tnedfracpi2$
$sin t=sqrtleft(1-dfrac1xright)^2=dfracsqrtx^2-1$ as $sin tge0$
$tan t=sin tsec t=?$
$$1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1=dfraccos^4t-8cos^2t+8-4sin tcos^2t+8sin tcos^4t$$
Now $cos^4t-8cos^2t+8-4sin tcos^2t+8sin t$
$=(1-sin^2t)^2-8(1-sin^2t)+8-4sin t(1-sin^2t)+8sin t$
$=sin^4t+4sin^3t+6sin^2t+4sin t+1$
$=(sin t+1)^4$
answered Aug 2 at 8:53
lab bhattacharjee
214k14152263
answered Aug 2 at 8:53
lab bhattacharjee
214k14152263
answered Aug 2 at 8:53
lab bhattacharjee
214k14152263
answered Aug 2 at 8:53
lab bhattacharjee
214k14152263
214k14152263
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You could always put it into WolframAlpha, see if you get anything remotely nice and if so reverse engineer a solution from there. But be warned: some times WA gives ugly answers when there exist nice ones.
– Arthur
Aug 1 at 4:33
WA doesn't give anything nice back: wolframalpha.com/input/…
– Mason
Aug 1 at 4:36
3
All of these kinds of integrals are jokes in bad taste. The universe is laughing at the limited expressive strength of our mathematics.
– Daniel
Aug 1 at 6:32
3
Can't the title be more descriptive? .........
– user202729
Aug 1 at 7:21
@Mason: WA often gives Wrong Answers, so even if it says something you still have to know how to verify it.
– user21820
Aug 1 at 7:40